entityMap|blocks|key|fekiv|text|df+=+pd.DataFrame({'Year':+['2014',+'2015'],+'quarter':+['q1',+'q2']})
df['period']+=+df[['Year',+'quarter']].apply(lambda+x:+''.join(x),+axis=1)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|aff4f|生成此数据帧|unstyled|72c0h|Year+quarter++period
0++2014++++++q1++2014q1
1++2015++++++q2++2015q2|7kmjk|此方法可泛化为任意数量的字符串列，方法是将|1t0sp|对于数据帧的任何列切片，例如|at1o5|..。|2n9p9|您可以查看有关apply()方法的更多信息|c0m8s|这里^0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|T|9|@]|A|@]|B|$C|D]]|$2|E|4|F|6|G|8|U|9|@]|A|@]|B|$]]|$2|H|4|I|6|7|8|V|9|@]|A|@]|B|$C|D]]|$2|J|4|K|6|G|8|W|9|@]|A|@]|B|$]]|$2|L|4|M|6|G|8|X|9|@]|A|@]|B|$]]|$2|N|4|O|6|G|8|Y|9|@]|A|@]|B|$]]|$2|P|4|Q|6|G|8|Z|9|@]|A|@]|B|$]]|$2|R|4|S|6|G|8|10|9|@]|A|@]|B|$]]]]

<pre><code>df = pd.DataFrame({'Year': ['2014', '2015'], 'quarter': ['q1', 'q2']})
df['period'] = df[['Year', 'quarter']].apply(lambda x: ''.join(x), axis=1)
</code></pre>

Yields this dataframe

<pre><code> Year quarter period
0 2014 q1 2014q1
1 2015 q2 2015q2
</code></pre>

This method generalizes to an arbitrary number of string columns by replacing <code>df[['Year', 'quarter']]</code> with any column slice of your dataframe, e.g. <code>df.iloc[:,0:2].apply(lambda x: ''.join(x), axis=1)</code>.

You can check more information about apply() method <a href="http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.apply.html" rel="noreferrer">here</a>

entityMap|blocks|key|bfjcb|text|虽然@silvado的答案是好的，如果你改变|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1q55a|至|2n6b|它会更快：|3f2hj|import+pandas+as+pd
df+=+pd.DataFrame({'Year':+['2014',+'2015'],+'quarter':+['q1',+'q2']})

In+[131]:+%25timeit+df["Year"].map(str)
10000+loops,+best+of+3:+132+us+per+loop

In+[132]:+%25timeit+df["Year"].astype(str)
10000+loops,+best+of+3:+82.2+us+per+loop|code-block|syntax|javascript^0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|L|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|M|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|N|9|@]|A|@]|B|$]]|$2|G|4|H|6|I|8|O|9|@]|A|@]|B|$J|K]]]]

Although the @silvado answer is good if you change <code>df.map(str)</code> to <code>df.astype(str)</code> it will be faster:

<pre><code>import pandas as pd
df = pd.DataFrame({'Year': ['2014', '2015'], 'quarter': ['q1', 'q2']})

In [131]: %timeit df["Year"].map(str)
10000 loops, best of 3: 132 us per loop

In [132]: %timeit df["Year"].astype(str)
10000 loops, best of 3: 82.2 us per loop
</code></pre>

entityMap|blocks|key|1n973|text|方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3hgru|的|c3cmb|访问者|dc7up|在这一点上，效果真的很好：|cc9t2|>>>+import+pandas+as+pd
>>>+df+=+pd.DataFrame([["2014",+"q1"],+
...++++++++++++++++++++["2015",+"q3"]],
...+++++++++++++++++++columns=('Year',+'Quarter'))
>>>+print(df)
+++Year+Quarter
0++2014++++++q1
1++2015++++++q3
>>>+df['Period']+=+df.Year.str.cat(df.Quarter)
>>>+print(df)
+++Year+Quarter++Period
0++2014++++++q1++2014q1
1++2015++++++q3++2015q3|code-block|syntax|javascript|e4584|甚至允许您添加分隔符，例如，假设您只有年份和期间的整数，您可以这样做：|2rs14|>>>+import+pandas+as+pd
>>>+df+=+pd.DataFrame([[2014,+1],
...++++++++++++++++++++[2015,+3]],
...+++++++++++++++++++columns=('Year',+'Quarter'))
>>>+print(df)
+++Year+Quarter
0++2014+++++++1
1++2015+++++++3
>>>+df['Period']+=+df.Year.astype(str).str.cat(df.Quarter.astype(str),+sep='q')
>>>+print(df)
+++Year+Quarter++Period
0++2014+++++++1++2014q1
1++2015+++++++3++2015q3|87edq|连接多个列只需将序列列表或包含除第一列之外的所有列的数据帧作为参数传递给|f3ncc|在第一列(系列)上调用：|ngu5|>>>+df+=+pd.DataFrame(
...+++++[['USA',+'Nevada',+'Las+Vegas'],
...++++++['Brazil',+'Pernambuco',+'Recife']],
...+++++columns=['Country',+'State',+'City'],
...+)
>>>+df['AllTogether']+=+df['Country'].str.cat(df[['State',+'City']],+sep='+-+')
>>>+print(df)
++Country+++++++State+++++++City+++++++++++++++++++AllTogether
0+++++USA++++++Nevada++Las+Vegas++++++USA+-+Nevada+-+Las+Vegas
1++Brazil++Pernambuco+++++Recife++Brazil+-+Pernambuco+-+Recife|e91i2|请注意，如果您的pandas+dataframe/series具有空值，则需要包括参数na|cctuj|_|aar75|表示将NaN值替换为字符串，否则组合列默认为NaN。^0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|13|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|14|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|15|9|@]|A|@]|B|$]]|$2|G|4|H|6|7|8|16|9|@]|A|@]|B|$]]|$2|I|4|J|6|K|8|17|9|@]|A|@]|B|$L|M]]|$2|N|4|O|6|7|8|18|9|@]|A|@]|B|$]]|$2|P|4|Q|6|K|8|19|9|@]|A|@]|B|$L|M]]|$2|R|4|S|6|7|8|1A|9|@]|A|@]|B|$]]|$2|T|4|U|6|7|8|1B|9|@]|A|@]|B|$]]|$2|V|4|W|6|K|8|1C|9|@]|A|@]|B|$L|M]]|$2|X|4|Y|6|7|8|1D|9|@]|A|@]|B|$]]|$2|Z|4|10|6|7|8|1E|9|@]|A|@]|B|$]]|$2|11|4|12|6|7|8|1F|9|@]|A|@]|B|$]]]]

The method <a href="http://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.str.cat.html#pandas.Series.str.cat" rel="noreferrer"><code>cat()</code> of the <code>.str</code> accessor</a> works really well for this:

<pre><code>&gt;&gt;&gt; import pandas as pd
&gt;&gt;&gt; df = pd.DataFrame([["2014", "q1"], 
... ["2015", "q3"]],
... columns=('Year', 'Quarter'))
&gt;&gt;&gt; print(df)
 Year Quarter
0 2014 q1
1 2015 q3
&gt;&gt;&gt; df['Period'] = df.Year.str.cat(df.Quarter)
&gt;&gt;&gt; print(df)
 Year Quarter Period
0 2014 q1 2014q1
1 2015 q3 2015q3
</code></pre>

<code>cat()</code> even allows you to add a separator so, for example, suppose you only have integers for year and period, you can do this:

<pre><code>&gt;&gt;&gt; import pandas as pd
&gt;&gt;&gt; df = pd.DataFrame([[2014, 1],
... [2015, 3]],
... columns=('Year', 'Quarter'))
&gt;&gt;&gt; print(df)
 Year Quarter
0 2014 1
1 2015 3
&gt;&gt;&gt; df['Period'] = df.Year.astype(str).str.cat(df.Quarter.astype(str), sep='q')
&gt;&gt;&gt; print(df)
 Year Quarter Period
0 2014 1 2014q1
1 2015 3 2015q3
</code></pre>

Joining multiple columns is just a matter of passing either a list of series or a dataframe containing all but the first column as a parameter to <code>str.cat()</code> invoked on the first column (Series):

<pre><code>&gt;&gt;&gt; df = pd.DataFrame(
... [['USA', 'Nevada', 'Las Vegas'],
... ['Brazil', 'Pernambuco', 'Recife']],
... columns=['Country', 'State', 'City'],
... )
&gt;&gt;&gt; df['AllTogether'] = df['Country'].str.cat(df[['State', 'City']], sep=' - ')
&gt;&gt;&gt; print(df)
 Country State City AllTogether
0 USA Nevada Las Vegas USA - Nevada - Las Vegas
1 Brazil Pernambuco Recife Brazil - Pernambuco - Recife
</code></pre>

Do note that if your pandas dataframe/series has null values, you need to include the parameter na_rep to replace the NaN values with a string, otherwise the combined column will default to NaN.

entityMap|blocks|key|5r9ma|text|使用lamba函数，这次使用string.format()。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|ph55|import+pandas+as+pd
df+=+pd.DataFrame({'Year':+['2014',+'2015'],+'Quarter':+['q1',+'q2']})
print+df
df['YearQuarter']+=+df[['Year','Quarter']].apply(lambda+x+:+'{}{}'.format(x[0],x[1]),+axis=1)
print+df

++Quarter++Year
0++++++q1++2014
1++++++q2++2015
++Quarter++Year+YearQuarter
0++++++q1++2014++++++2014q1
1++++++q2++2015++++++2015q2|code-block|syntax|javascript|6nv1u|这允许您使用非字符串并根据需要重新格式化值。|82rci|import+pandas+as+pd
df+=+pd.DataFrame({'Year':+['2014',+'2015'],+'Quarter':+[1,+2]})
print+df.dtypes
print+df

df['YearQuarter']+=+df[['Year','Quarter']].apply(lambda+x+:+'{}q{}'.format(x[0],x[1]),+axis=1)
print+df

Quarter+++++int64
Year+++++++object
dtype:+object
+++Quarter++Year
0++++++++1++2014
1++++++++2++2015
+++Quarter++Year+YearQuarter
0++++++++1++2014++++++2014q1
1++++++++2++2015++++++2015q2^0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|L|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|M|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|N|9|@]|A|@]|B|$]]|$2|J|4|K|6|E|8|O|9|@]|A|@]|B|$F|G]]]]

Use of a lamba function this time with string.format(). 

<pre><code>import pandas as pd
df = pd.DataFrame({'Year': ['2014', '2015'], 'Quarter': ['q1', 'q2']})
print df
df['YearQuarter'] = df[['Year','Quarter']].apply(lambda x : '{}{}'.format(x[0],x[1]), axis=1)
print df

 Quarter Year
0 q1 2014
1 q2 2015
 Quarter Year YearQuarter
0 q1 2014 2014q1
1 q2 2015 2015q2
</code></pre>

This allows you to work with non-strings and reformat values as needed.

<pre><code>import pandas as pd
df = pd.DataFrame({'Year': ['2014', '2015'], 'Quarter': [1, 2]})
print df.dtypes
print df

df['YearQuarter'] = df[['Year','Quarter']].apply(lambda x : '{}q{}'.format(x[0],x[1]), axis=1)
print df

Quarter int64
Year object
dtype: object
 Quarter Year
0 1 2014
1 2 2015
 Quarter Year YearQuarter
0 1 2014 2014q1
1 2 2015 2015q2
</code></pre>

entityMap|0|type|IMAGE|mutability|IMMUTABLE|data|imageUrl|https://ask.qcloudimg.com/http-save/actteam/d592ccde50dc7d8005d457994cf9f741.png|imageAlt|blocks|key|e5qqb|text|小数据集(<+150行)|unstyled|depth|inlineStyleRanges|entityRanges|4r3up|[''.join(i)+for+i+in+zip(df["Year"].map(str),df["quarter"])]|code-block|syntax|javascript|d537m|或者稍微慢一点，但更紧凑：|6fgb6|df.Year.str.cat(df.quarter)|4smbc|更大的数据集(>+150行)|2vta1|df['Year'].astype(str)+%2B+df['quarter']|v27a|更新：|edqgt|时序图Pandas+0.23.4|dthb4|​|d4tac|59bd7|?|atomic|offset|length|bvjf4|b077u|a0dj3|让我们在200K行DF上测试它：|8p5u1|In+[250]:+df
Out[250]:
+++Year+quarter
0++2014++++++q1
1++2015++++++q2

In+[251]:+df+=+pd.concat([df]+*+10**5)

In+[252]:+df.shape
Out[252]:+(200000,+2)|59fcd|a4094|使用Pandas+0.19.0的新计时|9rlsq|计时|bjn2h|不进行CPU/GPU优化(从最快到最慢排序)：|3rhv0|In+[107]:+%25timeit+df['Year'].astype(str)+%2B+df['quarter']
10+loops,+best+of+3:+131+ms+per+loop

In+[106]:+%25timeit+df['Year'].map(str)+%2B+df['quarter']
10+loops,+best+of+3:+161+ms+per+loop

In+[108]:+%25timeit+df.Year.str.cat(df.quarter)
10+loops,+best+of+3:+189+ms+per+loop

In+[109]:+%25timeit+df.loc[:,+['Year','quarter']].astype(str).sum(axis=1)
1+loop,+best+of+3:+567+ms+per+loop

In+[110]:+%25timeit+df[['Year','quarter']].astype(str).sum(axis=1)
1+loop,+best+of+3:+584+ms+per+loop

In+[111]:+%25timeit+df[['Year','quarter']].apply(lambda+x+:+'{}{}'.format(x[0],x[1]),+axis=1)
1+loop,+best+of+3:+24.7+s+per+loop|4e66r|948kq|使用CPU/GPU优化：|2dc4p|In+[113]:+%25timeit+df['Year'].astype(str)+%2B+df['quarter']
10+loops,+best+of+3:+53.3+ms+per+loop

In+[114]:+%25timeit+df['Year'].map(str)+%2B+df['quarter']
10+loops,+best+of+3:+65.5+ms+per+loop

In+[115]:+%25timeit+df.Year.str.cat(df.quarter)
10+loops,+best+of+3:+79.9+ms+per+loop

In+[116]:+%25timeit+df.loc[:,+['Year','quarter']].astype(str).sum(axis=1)
1+loop,+best+of+3:+230+ms+per+loop

In+[117]:+%25timeit+df[['Year','quarter']].astype(str).sum(axis=1)
1+loop,+best+of+3:+230+ms+per+loop

In+[118]:+%25timeit+df[['Year','quarter']].apply(lambda+x+:+'{}{}'.format(x[0],x[1]),+axis=1)
1+loop,+best+of+3:+9.38+s+per+loop|7cna1|回答贡献者|at67n|@anton-vbr^0|0|0|0|0|0|0|0|0|0|0|0|1|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|$1|$2|3|4|5|6|$7|8|9|-4]]]|A|@$B|C|D|E|2|F|G|1W|H|@]|I|@]|6|$]]|$B|J|D|K|2|L|G|1X|H|@]|I|@]|6|$M|N]]|$B|O|D|P|2|F|G|1Y|H|@]|I|@]|6|$]]|$B|Q|D|R|2|L|G|1Z|H|@]|I|@]|6|$M|N]]|$B|S|D|T|2|F|G|20|H|@]|I|@]|6|$]]|$B|U|D|V|2|L|G|21|H|@]|I|@]|6|$M|N]]|$B|W|D|X|2|F|G|22|H|@]|I|@]|6|$]]|$B|Y|D|Z|2|F|G|23|H|@]|I|@]|6|$]]|$B|10|D|11|2|F|G|24|H|@]|I|@]|6|$]]|$B|12|D|-4|2|F|G|25|H|@]|I|@]|6|$]]|$B|13|D|14|2|15|G|26|H|@]|I|@$16|27|17|28|B|29]]|6|$]]|$B|18|D|-4|2|F|G|2A|H|@]|I|@]|6|$]]|$B|19|D|11|2|F|G|2B|H|@]|I|@]|6|$]]|$B|1A|D|1B|2|F|G|2C|H|@]|I|@]|6|$]]|$B|1C|D|1D|2|L|G|2D|H|@]|I|@]|6|$M|N]]|$B|1E|D|X|2|F|G|2E|H|@]|I|@]|6|$]]|$B|1F|D|1G|2|F|G|2F|H|@]|I|@]|6|$]]|$B|1H|D|1I|2|F|G|2G|H|@]|I|@]|6|$]]|$B|1J|D|1K|2|F|G|2H|H|@]|I|@]|6|$]]|$B|1L|D|1M|2|L|G|2I|H|@]|I|@]|6|$M|N]]|$B|1N|D|1I|2|F|G|2J|H|@]|I|@]|6|$]]|$B|1O|D|1P|2|F|G|2K|H|@]|I|@]|6|$]]|$B|1Q|D|1R|2|L|G|2L|H|@]|I|@]|6|$M|N]]|$B|1S|D|1T|2|F|G|2M|H|@]|I|@]|6|$]]|$B|1U|D|1V|2|F|G|2N|H|@]|I|@]|6|$]]]]

<h2>Small data-sets (&lt; 150rows)</h2>
<pre><code>[''.join(i) for i in zip(df[&quot;Year&quot;].map(str),df[&quot;quarter&quot;])]
</code></pre>
or slightly slower but more compact:
<pre><code>df.Year.str.cat(df.quarter)
</code></pre>
<h2>Larger data sets (&gt; 150rows)</h2>
<pre><code>df['Year'].astype(str) + df['quarter']
</code></pre>
<hr />
UPDATE: Timing graph Pandas 0.23.4
<a href="https://i.stack.imgur.com/TSHfx.png" rel="noreferrer"><img src="https://i.stack.imgur.com/TSHfx.png" alt="enter image description here" /></a>
Let's test it on 200K rows DF:
<pre><code>In [250]: df
Out[250]:
 Year quarter
0 2014 q1
1 2015 q2

In [251]: df = pd.concat([df] * 10**5)

In [252]: df.shape
Out[252]: (200000, 2)
</code></pre>
UPDATE: new timings using Pandas 0.19.0
Timing without CPU/GPU optimization (sorted from fastest to slowest):
<pre><code>In [107]: %timeit df['Year'].astype(str) + df['quarter']
10 loops, best of 3: 131 ms per loop

In [106]: %timeit df['Year'].map(str) + df['quarter']
10 loops, best of 3: 161 ms per loop

In [108]: %timeit df.Year.str.cat(df.quarter)
10 loops, best of 3: 189 ms per loop

In [109]: %timeit df.loc[:, ['Year','quarter']].astype(str).sum(axis=1)
1 loop, best of 3: 567 ms per loop

In [110]: %timeit df[['Year','quarter']].astype(str).sum(axis=1)
1 loop, best of 3: 584 ms per loop

In [111]: %timeit df[['Year','quarter']].apply(lambda x : '{}{}'.format(x[0],x[1]), axis=1)
1 loop, best of 3: 24.7 s per loop
</code></pre>
Timing using CPU/GPU optimization:
<pre><code>In [113]: %timeit df['Year'].astype(str) + df['quarter']
10 loops, best of 3: 53.3 ms per loop

In [114]: %timeit df['Year'].map(str) + df['quarter']
10 loops, best of 3: 65.5 ms per loop

In [115]: %timeit df.Year.str.cat(df.quarter)
10 loops, best of 3: 79.9 ms per loop

In [116]: %timeit df.loc[:, ['Year','quarter']].astype(str).sum(axis=1)
1 loop, best of 3: 230 ms per loop

In [117]: %timeit df[['Year','quarter']].astype(str).sum(axis=1)
1 loop, best of 3: 230 ms per loop

In [118]: %timeit df[['Year','quarter']].apply(lambda x : '{}{}'.format(x[0],x[1]), axis=1)
1 loop, best of 3: 9.38 s per loop
</code></pre>
Answer contribution by <a href="https://stackoverflow.com/users/7386332/anton-vbr">@anton-vbr</a>

entityMap|blocks|key|ao3d|text|下面是一个我觉得非常通用的实现：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|cmi9c|In+[1]:+import+pandas+as+pd+

In+[2]:+df+=+pd.DataFrame([[0,+'the',+'quick',+'brown'],
+++...:++++++++++++++++++++[1,+'fox',+'jumps',+'over'],+
+++...:++++++++++++++++++++[2,+'the',+'lazy',+'dog']],
+++...:+++++++++++++++++++columns=['c0',+'c1',+'c2',+'c3'])

In+[3]:+def+str_join(df,+sep,+*cols):
+++...:+++++from+functools+import+reduce
+++...:+++++return+reduce(lambda+x,+y:+x.astype(str).str.cat(y.astype(str),+sep=sep),+
+++...:+++++++++++++++++++[df[col]+for+col+in+cols])
+++...:+

In+[4]:+df['cat']+=+str_join(df,+'-',+'c0',+'c1',+'c2',+'c3')

In+[5]:+df
Out[5]:+
+++c0+++c1+++++c2+++++c3++++++++++++++++cat
0+++0++the++quick++brown++0-the-quick-brown
1+++1++fox++jumps+++over+++1-fox-jumps-over
2+++2++the+++lazy++++dog+++++2-the-lazy-dog|code-block|syntax|javascript^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|I|9|@]|A|@]|B|$F|G]]]]

Here is an implementation that I find very versatile: 

<pre><code>In [1]: import pandas as pd 

In [2]: df = pd.DataFrame([[0, 'the', 'quick', 'brown'],
 ...: [1, 'fox', 'jumps', 'over'], 
 ...: [2, 'the', 'lazy', 'dog']],
 ...: columns=['c0', 'c1', 'c2', 'c3'])

In [3]: def str_join(df, sep, *cols):
 ...: from functools import reduce
 ...: return reduce(lambda x, y: x.astype(str).str.cat(y.astype(str), sep=sep), 
 ...: [df[col] for col in cols])
 ...: 

In [4]: df['cat'] = str_join(df, '-', 'c0', 'c1', 'c2', 'c3')

In [5]: df
Out[5]: 
 c0 c1 c2 c3 cat
0 0 the quick brown 0-the-quick-brown
1 1 fox jumps over 1-fox-jumps-over
2 2 the lazy dog 2-the-lazy-dog
</code></pre>

entityMap|blocks|key|3omha|text|def+madd(x):
++++"""Performs+element-wise+string+concatenation+with+multiple+input+arrays.

++++Args:
++++++++x:+iterable+of+np.array.

++++Returns:+np.array.
++++"""
++++for+i,+arr+in+enumerate(x):
++++++++if+type(arr.item(0))+is+not+str:
++++++++++++x[i]+=+x[i].astype(str)
++++return+reduce(np.core.defchararray.add,+x)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|6jjo3|例如：|unstyled|6srhn|data+=+list(zip([2000]*4,+['q1',+'q2',+'q3',+'q4']))
df+=+pd.DataFrame(data=data,+columns=['Year',+'quarter'])
df['period']+=+madd([df[col].values+for+col+in+['Year',+'quarter']])

df

++++Year++++quarter+period
0+++2000++++q1++2000q1
1+++2000++++q2++2000q2
2+++2000++++q3++2000q3
3+++2000++++q4++2000q4^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|J|9|@]|A|@]|B|$C|D]]|$2|E|4|F|6|G|8|K|9|@]|A|@]|B|$]]|$2|H|4|I|6|7|8|L|9|@]|A|@]|B|$C|D]]]]

<pre><code>def madd(x):
 """Performs element-wise string concatenation with multiple input arrays.

 Args:
 x: iterable of np.array.

 Returns: np.array.
 """
 for i, arr in enumerate(x):
 if type(arr.item(0)) is not str:
 x[i] = x[i].astype(str)
 return reduce(np.core.defchararray.add, x)
</code></pre>

For example:

<pre><code>data = list(zip([2000]*4, ['q1', 'q2', 'q3', 'q4']))
df = pd.DataFrame(data=data, columns=['Year', 'quarter'])
df['period'] = madd([df[col].values for col in ['Year', 'quarter']])

df

 Year quarter period
0 2000 q1 2000q1
1 2000 q2 2000q2
2 2000 q3 2000q3
3 2000 q4 2000q4
</code></pre>

entityMap|blocks|key|cvdqn|text|正如前面提到的，您必须将每个列转换为字符串，然后使用加号运算符来组合两个字符串列。通过使用NumPy，您可以获得较大的性能改进。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|b5i2o|%25timeit+df['Year'].values.astype(str)+%2B+df.quarter
71.1+ms+±+3.76+ms+per+loop+(mean+±+std.+dev.+of+7+runs,+10+loops+each)

%25timeit+df['Year'].astype(str)+%2B+df['quarter']
565+ms+±+22.3+ms+per+loop+(mean+±+std.+dev.+of+7+runs,+1+loop+each)|code-block|syntax|javascript^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|I|9|@]|A|@]|B|$F|G]]]]

As many have mentioned previously, you must convert each column to string and then use the plus operator to combine two string columns. You can get a large performance improvement by using NumPy.

<pre><code>%timeit df['Year'].values.astype(str) + df.quarter
71.1 ms ± 3.76 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

%timeit df['Year'].astype(str) + df['quarter']
565 ms ± 22.3 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
</code></pre>

entityMap|blocks|key|35jq3|text|效率更高的是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|7u1tq|def+concat_df_str1(df):
++++"""+run+time:+1.3416s+"""
++++return+pd.Series([''.join(row.astype(str))+for+row+in+df.values],+index=df.index)|code-block|syntax|javascript|6b5br|这是一个时间测试：|mhte|import+numpy+as+np
import+pandas+as+pd

from+time+import+time


def+concat_df_str1(df):
++++"""+run+time:+1.3416s+"""
++++return+pd.Series([''.join(row.astype(str))+for+row+in+df.values],+index=df.index)


def+concat_df_str2(df):
++++"""+run+time:+5.2758s+"""
++++return+df.astype(str).sum(axis=1)


def+concat_df_str3(df):
++++"""+run+time:+5.0076s+"""
++++df+=+df.astype(str)
++++return+df[0]+%2B+df[1]+%2B+df[2]+%2B+df[3]+%2B+df[4]+%2B+\
+++++++++++df[5]+%2B+df[6]+%2B+df[7]+%2B+df[8]+%2B+df[9]


def+concat_df_str4(df):
++++"""+run+time:+7.8624s+"""
++++return+df.astype(str).apply(lambda+x:+''.join(x),+axis=1)


def+main():
++++df+=+pd.DataFrame(np.zeros(1000000).reshape(100000,+10))
++++df+=+df.astype(int)

++++time1+=+time()
++++df_en+=+concat_df_str4(df)
++++print('run+time:+%25.4fs'+%25+(time()+-+time1))
++++print(df_en.head(10))


if+__name__+==+'__main__':
++++main()|csrll|最终，当|bfvo8|(合并|8p37h|_|54th4|df|c19nj|jv2i|str2)，则结果不是简单的连接，它将转换为整数。^0|0|0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|W|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|X|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|Y|9|@]|A|@]|B|$]]|$2|J|4|K|6|E|8|Z|9|@]|A|@]|B|$F|G]]|$2|L|4|M|6|7|8|10|9|@]|A|@]|B|$]]|$2|N|4|O|6|7|8|11|9|@]|A|@]|B|$]]|$2|P|4|Q|6|7|8|12|9|@]|A|@]|B|$]]|$2|R|4|S|6|7|8|13|9|@]|A|@]|B|$]]|$2|T|4|Q|6|7|8|14|9|@]|A|@]|B|$]]|$2|U|4|V|6|7|8|15|9|@]|A|@]|B|$]]]]

more efficient is

<pre><code>def concat_df_str1(df):
 """ run time: 1.3416s """
 return pd.Series([''.join(row.astype(str)) for row in df.values], index=df.index)
</code></pre>

and here is a time test:

<pre><code>import numpy as np
import pandas as pd

from time import time


def concat_df_str1(df):
 """ run time: 1.3416s """
 return pd.Series([''.join(row.astype(str)) for row in df.values], index=df.index)


def concat_df_str2(df):
 """ run time: 5.2758s """
 return df.astype(str).sum(axis=1)


def concat_df_str3(df):
 """ run time: 5.0076s """
 df = df.astype(str)
 return df[0] + df[1] + df[2] + df[3] + df[4] + \
 df[5] + df[6] + df[7] + df[8] + df[9]


def concat_df_str4(df):
 """ run time: 7.8624s """
 return df.astype(str).apply(lambda x: ''.join(x), axis=1)


def main():
 df = pd.DataFrame(np.zeros(1000000).reshape(100000, 10))
 df = df.astype(int)

 time1 = time()
 df_en = concat_df_str4(df)
 print('run time: %.4fs' % (time() - time1))
 print(df_en.head(10))


if __name__ == '__main__':
 main()
</code></pre>

final, when <code>sum</code>(concat_df_str2) is used, the result is not simply concat, it will trans to integer.

entityMap|blocks|key|b5tje|text|使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1t2an|..。|80a0d|df['Period']+=+df['Year'].combine_first(df['Quarter'])|code-block|syntax|javascript^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|J|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|K|9|@]|A|@]|B|$]]|$2|E|4|F|6|G|8|L|9|@]|A|@]|B|$H|I]]]]

Use <code>.combine_first</code>.

<pre><code>df['Period'] = df['Year'].combine_first(df['Quarter'])
</code></pre>

entityMap|0|type|IMAGE|mutability|IMMUTABLE|data|imageUrl|https://ask.qcloudimg.com/http-save/actteam/b770075d5113d3b3447e339cd827c70d.png|imageAlt|blocks|key|cv5r|text|使用|unstyled|depth|inlineStyleRanges|entityRanges|2aqkh|可能会更快：|b45fk|df["period"]+=+[''.join(i)+for+i+in+zip(df["Year"].map(str),df["quarter"])]|code-block|syntax|javascript|2q1h|图表：|68i14|​|c0av5|6rl3|?|atomic|offset|length|f9rj8|5rumf|e1on3|import+pandas+as+pd
import+numpy+as+np
import+timeit
import+matplotlib.pyplot+as+plt
from+collections+import+defaultdict

df+=+pd.DataFrame({'Year':+['2014',+'2015'],+'quarter':+['q1',+'q2']})

myfuncs+=+{
"df['Year'].astype(str)+%2B+df['quarter']":
++++lambda:+df['Year'].astype(str)+%2B+df['quarter'],
"df['Year'].map(str)+%2B+df['quarter']":
++++lambda:+df['Year'].map(str)+%2B+df['quarter'],
"df.Year.str.cat(df.quarter)":
++++lambda:+df.Year.str.cat(df.quarter),
"df.loc[:,+['Year','quarter']].astype(str).sum(axis=1)":
++++lambda:+df.loc[:,+['Year','quarter']].astype(str).sum(axis=1),
"df[['Year','quarter']].astype(str).sum(axis=1)":
++++lambda:+df[['Year','quarter']].astype(str).sum(axis=1),
++++"df[['Year','quarter']].apply(lambda+x+:+'{}{}'.format(x[0],x[1]),+axis=1)":
++++lambda:+df[['Year','quarter']].apply(lambda+x+:+'{}{}'.format(x[0],x[1]),+axis=1),
++++"[''.join(i)+for+i+in+zip(dataframe['Year'].map(str),dataframe['quarter'])]":
++++lambda:+[''.join(i)+for+i+in+zip(df["Year"].map(str),df["quarter"])]
}

d+=+defaultdict(dict)
step+=+10
cont+=+True
while+cont:
++++lendf+=+len(df);+print(lendf)
++++for+k,v+in+myfuncs.items():
++++++++iters+=+1
++++++++t+=+0
++++++++while+t+<+0.2:
++++++++++++ts+=+timeit.repeat(v,+number=iters,+repeat=3)
++++++++++++t+=+min(ts)
++++++++++++iters+*=+10
++++++++d[k][lendf]+=+t/iters
++++++++if+t+>+2:+cont+=+False
++++df+=+pd.concat([df]*step)

pd.DataFrame(d).plot().legend(loc='upper+center',+bbox_to_anchor=(0.5,+-0.15))
plt.yscale('log');+plt.xscale('log');+plt.ylabel('seconds');+plt.xlabel('df+rows')
plt.show()^0|0|0|0|0|0|0|0|1|0|0|0|0^^$0|$1|$2|3|4|5|6|$7|8|9|-4]]]|A|@$B|C|D|E|2|F|G|14|H|@]|I|@]|6|$]]|$B|J|D|K|2|F|G|15|H|@]|I|@]|6|$]]|$B|L|D|M|2|N|G|16|H|@]|I|@]|6|$O|P]]|$B|Q|D|R|2|F|G|17|H|@]|I|@]|6|$]]|$B|S|D|T|2|F|G|18|H|@]|I|@]|6|$]]|$B|U|D|-4|2|F|G|19|H|@]|I|@]|6|$]]|$B|V|D|W|2|X|G|1A|H|@]|I|@$Y|1B|Z|1C|B|1D]]|6|$]]|$B|10|D|-4|2|F|G|1E|H|@]|I|@]|6|$]]|$B|11|D|T|2|F|G|1F|H|@]|I|@]|6|$]]|$B|12|D|13|2|N|G|1G|H|@]|I|@]|6|$O|P]]]]

Using <code>zip</code> could be even quicker:

<pre><code>df["period"] = [''.join(i) for i in zip(df["Year"].map(str),df["quarter"])]
</code></pre>

Graph:

<a href="https://i.stack.imgur.com/QaLDM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QaLDM.png" alt="enter image description here"></a>

<pre><code>import pandas as pd
import numpy as np
import timeit
import matplotlib.pyplot as plt
from collections import defaultdict

df = pd.DataFrame({'Year': ['2014', '2015'], 'quarter': ['q1', 'q2']})

myfuncs = {
"df['Year'].astype(str) + df['quarter']":
 lambda: df['Year'].astype(str) + df['quarter'],
"df['Year'].map(str) + df['quarter']":
 lambda: df['Year'].map(str) + df['quarter'],
"df.Year.str.cat(df.quarter)":
 lambda: df.Year.str.cat(df.quarter),
"df.loc[:, ['Year','quarter']].astype(str).sum(axis=1)":
 lambda: df.loc[:, ['Year','quarter']].astype(str).sum(axis=1),
"df[['Year','quarter']].astype(str).sum(axis=1)":
 lambda: df[['Year','quarter']].astype(str).sum(axis=1),
 "df[['Year','quarter']].apply(lambda x : '{}{}'.format(x[0],x[1]), axis=1)":
 lambda: df[['Year','quarter']].apply(lambda x : '{}{}'.format(x[0],x[1]), axis=1),
 "[''.join(i) for i in zip(dataframe['Year'].map(str),dataframe['quarter'])]":
 lambda: [''.join(i) for i in zip(df["Year"].map(str),df["quarter"])]
}

d = defaultdict(dict)
step = 10
cont = True
while cont:
 lendf = len(df); print(lendf)
 for k,v in myfuncs.items():
 iters = 1
 t = 0
 while t &lt; 0.2:
 ts = timeit.repeat(v, number=iters, repeat=3)
 t = min(ts)
 iters *= 10
 d[k][lendf] = t/iters
 if t &gt; 2: cont = False
 df = pd.concat([df]*step)

pd.DataFrame(d).plot().legend(loc='upper center', bbox_to_anchor=(0.5, -0.15))
plt.yscale('log'); plt.xscale('log'); plt.ylabel('seconds'); plt.xlabel('df rows')
plt.show()
</code></pre>

entityMap|blocks|key|bq9dh|text|让我们假设您的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|dj8pp|是|e7mac|包含列|2bqc3|和|52bok|..。|dlkvg|import+pandas+as+pd
df+=+pd.DataFrame({'Quarter':'q1+q2+q3+q4'.split(),+'Year':'2000'})|code-block|syntax|javascript|d4fvk|假设我们想要查看数据帧；|1r6d7|df
>>>++Quarter++++Year
+++0++++q1++++++2000
+++1++++q2++++++2000
+++2++++q3++++++2000
+++3++++q4++++++2000|clthv|最后，将|f4dbv|e04bu|如下所示。|eugpk|df['Period']+=+df['Year']+%2B+'+'+%2B+df['Quarter']|fhlmt|您现在可以|c7f8j|8sbnj|来查看结果数据帧。|b8pvh|df
>>>++Quarter++++Year++++Period
++++0+++q1++++++2000++++2000+q1
++++1+++q2++++++2000++++2000+q2
++++2+++q3++++++2000++++2000+q3
++++3+++q4++++++2000++++2000+q4|41obr|如果您不想要年份和季度之间的空间，只需通过执行以下操作来删除它；|vv38|df['Period']+=+df['Year']+%2B+df['Quarter']^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|1B|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|1C|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|1D|9|@]|A|@]|B|$]]|$2|G|4|H|6|7|8|1E|9|@]|A|@]|B|$]]|$2|I|4|J|6|7|8|1F|9|@]|A|@]|B|$]]|$2|K|4|L|6|M|8|1G|9|@]|A|@]|B|$N|O]]|$2|P|4|Q|6|7|8|1H|9|@]|A|@]|B|$]]|$2|R|4|S|6|M|8|1I|9|@]|A|@]|B|$N|O]]|$2|T|4|U|6|7|8|1J|9|@]|A|@]|B|$]]|$2|V|4|H|6|7|8|1K|9|@]|A|@]|B|$]]|$2|W|4|X|6|7|8|1L|9|@]|A|@]|B|$]]|$2|Y|4|Z|6|M|8|1M|9|@]|A|@]|B|$N|O]]|$2|10|4|11|6|7|8|1N|9|@]|A|@]|B|$]]|$2|12|4|-4|6|7|8|1O|9|@]|A|@]|B|$]]|$2|13|4|14|6|7|8|1P|9|@]|A|@]|B|$]]|$2|15|4|16|6|M|8|1Q|9|@]|A|@]|B|$N|O]]|$2|17|4|18|6|7|8|1R|9|@]|A|@]|B|$]]|$2|19|4|1A|6|M|8|1S|9|@]|A|@]|B|$N|O]]]]

Let us suppose your <code>dataframe</code> is <code>df</code> with columns <code>Year</code> and <code>Quarter</code>.

<pre><code>import pandas as pd
df = pd.DataFrame({'Quarter':'q1 q2 q3 q4'.split(), 'Year':'2000'})
</code></pre>

Suppose we want to see the dataframe;

<pre><code>df
&gt;&gt;&gt; Quarter Year
 0 q1 2000
 1 q2 2000
 2 q3 2000
 3 q4 2000
</code></pre>

Finally, concatenate the <code>Year</code> and the <code>Quarter</code> as follows.

<pre><code>df['Period'] = df['Year'] + ' ' + df['Quarter']
</code></pre>

You can now <code>print</code> <code>df</code> to see the resulting dataframe.

<pre><code>df
&gt;&gt;&gt; Quarter Year Period
 0 q1 2000 2000 q1
 1 q2 2000 2000 q2
 2 q3 2000 2000 q3
 3 q4 2000 2000 q4
</code></pre>

If you do not want the space between the year and quarter, simply remove it by doing;

<pre><code>df['Period'] = df['Year'] + df['Quarter']
</code></pre>

entityMap|blocks|key|8jmuj|text|一个人可以使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|93b5u|分配|4501p|方法|90m9n|DataFrame|an30n|df=+(pd.DataFrame({'Year':+['2014',+'2015'],+'quarter':+['q1',+'q2']}).
++assign(period=lambda+x:+x.Year%2Bx.quarter+))|code-block|syntax|javascript^0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|N|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|O|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|P|9|@]|A|@]|B|$]]|$2|G|4|H|6|7|8|Q|9|@]|A|@]|B|$]]|$2|I|4|J|6|K|8|R|9|@]|A|@]|B|$L|M]]]]

One can use assign method of DataFrame:

<pre><code>df= (pd.DataFrame({'Year': ['2014', '2015'], 'quarter': ['q1', 'q2']}).
 assign(period=lambda x: x.Year+x.quarter ))
</code></pre>

entityMap|blocks|key|f6crq|text|此解决方案使用中间步骤|type|unstyled|depth|inlineStyleRanges|entityRanges|data|8f4ic|将DataFrame的两列压缩为包含列表的单个列|1k9h0|价值。这不仅适用于字符串，而且适用于所有类型的列数据类型|5ol7b|import+pandas+as+pd
df+=+pd.DataFrame({'Year':+['2014',+'2015'],+'quarter':+['q1',+'q2']})
df['list']=df[['Year','quarter']].values.tolist()
df['period']=df['list'].apply(''.join)
print(df)|code-block|syntax|javascript|1dfj5|结果：|8u2r5|Year+quarter++++++++list++period
0++2014++++++q1++[2014,+q1]++2014q1
1++2015++++++q2++[2015,+q2]++2015q2^0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|P|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|Q|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|R|9|@]|A|@]|B|$]]|$2|G|4|H|6|I|8|S|9|@]|A|@]|B|$J|K]]|$2|L|4|M|6|7|8|T|9|@]|A|@]|B|$]]|$2|N|4|O|6|I|8|U|9|@]|A|@]|B|$J|K]]]]

This solution uses an intermediate step compressing two columns of the DataFrame to a single column containing a list of the values.
This works not only for strings but for all kind of column-dtypes

<pre><code>import pandas as pd
df = pd.DataFrame({'Year': ['2014', '2015'], 'quarter': ['q1', 'q2']})
df['list']=df[['Year','quarter']].values.tolist()
df['period']=df['list'].apply(''.join)
print(df)
</code></pre>

Result:

<pre><code> Year quarter list period
0 2014 q1 [2014, q1] 2014q1
1 2015 q2 [2015, q2] 2015q2
</code></pre>

entityMap|blocks|key|8oi6n|text|下面是我对上述解决方案的总结，这些解决方案使用列的值之间的分隔符将两个具有int和str值的列连接/组合成一个新列。有三种解决方案可以实现这一目的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|433sk|#+be+cautious+about+the+separator,+some+symbols+may+cause+"SyntaxError:+EOL+while+scanning+string+literal".
#+e.g.+";;"+as+separator+would+raise+the+SyntaxError

separator+=+"&&"+

#+pd.Series.str.cat()+method+does+not+work+to+concatenate+/+combine+two+columns+with+int+value+and+str+value.+This+would+raise+"AttributeError:+Can+only+use+.cat+accessor+with+a+'category'+dtype"

df["period"]+=+df["Year"].map(str)+%2B+separator+%2B+df["quarter"]
df["period"]+=+df[['Year','quarter']].apply(lambda+x+:+'{}+&&+{}'.format(x[0],x[1]),+axis=1)
df["period"]+=+df.apply(lambda+x:+f'{x["Year"]}+&&+{x["quarter"]}',+axis=1)|code-block|syntax|javascript^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|I|9|@]|A|@]|B|$F|G]]]]

Here is my summary of the above solutions to concatenate / combine two columns with int and str value into a new column, using a separator between the values of columns. Three solutions work for this purpose.

<pre><code># be cautious about the separator, some symbols may cause "SyntaxError: EOL while scanning string literal".
# e.g. ";;" as separator would raise the SyntaxError

separator = "&amp;&amp;" 

# pd.Series.str.cat() method does not work to concatenate / combine two columns with int value and str value. This would raise "AttributeError: Can only use .cat accessor with a 'category' dtype"

df["period"] = df["Year"].map(str) + separator + df["quarter"]
df["period"] = df[['Year','quarter']].apply(lambda x : '{} &amp;&amp; {}'.format(x[0],x[1]), axis=1)
df["period"] = df.apply(lambda x: f'{x["Year"]} &amp;&amp; {x["quarter"]}', axis=1)
</code></pre>

entityMap|blocks|key|23skk|text|概括到多个列，为什么不：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|9np7k|columns+=+['whatever',+'columns',+'you',+'choose']
df['period']+=+df[columns].astype(str).sum(axis=1)|code-block|syntax|javascript^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|I|9|@]|A|@]|B|$F|G]]]]

generalising to multiple columns, why not:

<pre><code>columns = ['whatever', 'columns', 'you', 'choose']
df['period'] = df[columns].astype(str).sum(axis=1)
</code></pre>

entityMap|blocks|key|dr78f|text|我的观点..。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5g4ha|listofcols+=+['col1','col2','col3']
df['combined_cols']+=+''

for+column+in+listofcols:
++++df['combined_cols']+=+df['combined_cols']+%2B+'+'+%2B+df[column]
'''|code-block|syntax|javascript^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|I|9|@]|A|@]|B|$F|G]]]]

my take....
<pre><code>listofcols = ['col1','col2','col3']
df['combined_cols'] = ''

for column in listofcols:
 df['combined_cols'] = df['combined_cols'] + ' ' + df[column]
'''
</code></pre>

entityMap|blocks|key|6ta06|text|您可以使用Lambda：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|c5voo|combine_lambda+=+lambda+x:+'{}{}'.format(x.Year,+x.quater)|code-block|syntax|javascript|fvn8t|然后在创建新列时使用它：|6d4v3|df["period"]+=+df.apply(combine_lambda,+axis+=+1)^0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|L|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|M|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|N|9|@]|A|@]|B|$]]|$2|J|4|K|6|E|8|O|9|@]|A|@]|B|$F|G]]]]

You can use lambda:
<pre><code>combine_lambda = lambda x: '{}{}'.format(x.Year, x.quarter)
</code></pre>
And then use it with creating the new column:
<pre><code>df['period'] = df.apply(combine_lambda, axis = 1)
</code></pre>

I have a 20 x 4000 dataframe in Python using pandas. Two of these columns are named <code>Year</code> and <code>quarter</code>. I'd like to create a variable called <code>period</code> that makes <code>Year = 2000</code> and <code>quarter= q2</code> into <code>2000q2</code>.

Can anyone help with that?

Combine two columns of text in pandas dataframe

Python

 我有一个使用pandas的Python格式的20x4000数据帧。这些列中有两列被命名为 和 ..。我想创建一个名为 这使得 和 变成 ..。 有人能帮上忙吗？

问在pandas数据帧中组合两列文本
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