blocks|key|5865870|text|尝试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5865871|#Firstly+fill+NaN's+of+the+columns:
df[['Zone','Store','Department']]=df[['Zone','Store','Department']].fillna('')
#Finally:
df['id']=(df['Zone']%2B'.'%2Bdf['Store']%2B'.'%2Bdf['Department']).str.rstrip('.')|code-block|syntax|javascript|5865872|或|offset|length|style|BOLD|5865873|如果有超过4列，则使用apply()(性能方面，第一种方法比apply更快)：|CODE|5865874|#Firstly+fill+NaN's+of+the+columns:
df[['Zone','Store','Department']]=df[['Zone','Store','Department']].fillna('')
#Finally:
df['id']+=+df[['Zone','Store','Department']].apply('.'.join,+axis=1).str.rstrip('.')|5865875|entityMap^0|0|0|0|1|0|B|7|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|U|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|V|8|@$I|W|J|X|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|Y|8|@$I|Z|J|10|K|O]]|9|@]|A|$]]|$1|P|3|Q|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|R|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|S|$]]

try:
<pre><code>#Firstly fill NaN's of the columns:
df[['Zone','Store','Department']]=df[['Zone','Store','Department']].fillna('')
#Finally:
df['id']=(df['Zone']+'.'+df['Store']+'.'+df['Department']).str.rstrip('.')
</code></pre>
OR
If there are more then 4 columns then use <code>apply()</code>(performance wise the 1st method is fast as compared to apply):
<pre><code>#Firstly fill NaN's of the columns:
df[['Zone','Store','Department']]=df[['Zone','Store','Department']].fillna('')
#Finally:
df['id'] = df[['Zone','Store','Department']].apply('.'.join, axis=1).str.rstrip('.')
</code></pre>

blocks|key|4397432|text|你可以在这里使用带有str.join的df.agg。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4397433|df+=+df.fillna('')
df['id']+=+df[['Zone','Store','Department']].agg('.'.join,+axis=1)|code-block|syntax|javascript|4397434|entityMap^0|A|8|J|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

You can use <code>df.agg</code> with <code>str.join</code> here.
<pre><code>df = df.fillna('')
df['id'] = df[['Zone','Store','Department']].agg('.'.join, axis=1)
</code></pre>

blocks|key|4397481|text|这是一个泛化的例子。首先创建一个简单的小数据帧。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4397482|++++df+=+pd.DataFrame({'Zone':['a',+'b',+'c'],+'Store':['',+'st2',+'st3']})|code-block|syntax|javascript|4397483|然后创建名为'id‘的新列。|4397484|++++df['id']+=+(df.Zone+%2B+'.'+%2B+df.Store).str.strip('.')|4397485|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Here's like a generalized example.
First create a small simple dataframe.
<pre><code> df = pd.DataFrame({'Zone':['a', 'b', 'c'], 'Store':['', 'st2', 'st3']})
</code></pre>
Then create new column with name 'id'.
<pre><code> df['id'] = (df.Zone + '.' + df.Store).str.strip('.')
</code></pre>

blocks|key|2081241|text|可能工作过度了，但是这里有另一种使用reduce来解决这个问题的方法|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2081242|from+functools+import+reduce

cols+=+['Zone','Store','Department']
f+=+lambda+x,y+:+(x+%2B'.'%2By).str.rstrip(".")
#or#+f+=+lambda+x,y+:+x.str.cat(y,sep='.').str.rstrip(".")

df['id']+=+reduce(f,map(df.fillna('').get,+cols))|code-block|syntax|javascript|2081243|2081244|print(df)

++Zone+Store+Department++TTLSales++++++++++++++++id
0++APV+++NaN++++++++NaN+++++++220+++++++++++++++APV
1++APV++ST12++++++++NaN+++++++100++++++++++APV.ST12
2++APV++ST12+++++++Elec++++++++40+++++APV.ST12.Elec
3++APV++ST12++++Grocery++++++++20++APV.ST12.Grocery
4++APV++ST12++++++++CPG++++++++40++++++APV.ST12.CPG|2081245|entityMap^0|I|6|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|S|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|T|8|@]|D|@]|E|$]]|$1|L|3|M|5|H|7|U|8|@]|D|@]|E|$I|J]]|$1|N|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|O|$]]

Probably overworked , but here is another way of solving this using <code>reduce</code>:
<pre><code>from functools import reduce

cols = ['Zone','Store','Department']
f = lambda x,y : (x +'.'+y).str.rstrip(&quot;.&quot;)
#or# f = lambda x,y : x.str.cat(y,sep='.').str.rstrip(&quot;.&quot;)

df['id'] = reduce(f,map(df.fillna('').get, cols))
</code></pre>
<hr />
<pre><code>print(df)

 Zone Store Department TTLSales id
0 APV NaN NaN 220 APV
1 APV ST12 NaN 100 APV.ST12
2 APV ST12 Elec 40 APV.ST12.Elec
3 APV ST12 Grocery 20 APV.ST12.Grocery
4 APV ST12 CPG 40 APV.ST12.CPG
</code></pre>

blocks|key|4397537|text|使用以下代码|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4397538|df["id"]=df["Zone"]%2B"."%2Bdf["Store"]%2B"."%2Bdf["Department"]
df["id"]=df["id"].apply(lambda+x:x.rstrip("."))
df|code-block|syntax|javascript|4397539|结果|4397540|++++Zone++++Store+++Department++TTLSales++++id
0+++APV++++++++++++++++++++++++++220+++++++APV
1+++APV+++++ST12+++++++++++++++++100+++++++APV.ST12
2+++APV+++++ST12++++Elec+++++++++40++++++++APV.ST12.Elec
3+++APV+++++ST12++++Grocery++++++20++++++++APV.ST12.Grocery
4+++APV+++++ST12++++CPG++++++++++10++++++++APV.ST12.CPG|4397541|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

use the below code
<pre><code>df[&quot;id&quot;]=df[&quot;Zone&quot;]+&quot;.&quot;+df[&quot;Store&quot;]+&quot;.&quot;+df[&quot;Department&quot;]
df[&quot;id&quot;]=df[&quot;id&quot;].apply(lambda x:x.rstrip(&quot;.&quot;))
df
</code></pre>
result
<pre><code> Zone Store Department TTLSales id
0 APV 220 APV
1 APV ST12 100 APV.ST12
2 APV ST12 Elec 40 APV.ST12.Elec
3 APV ST12 Grocery 20 APV.ST12.Grocery
4 APV ST12 CPG 10 APV.ST12.CPG
</code></pre>

I have the following data frame:
<pre><code> Zone Store Department TTLSales 
0 APV 220 
1 APV ST12 100 
2 APV ST12 Elec 40 
3 APV ST12 Grocery 20 
4 APV ST12 CPG 40 
</code></pre>
I wish to include one column concatenating the values as:
<pre><code> Zone Store Department TTLSales id
0 APV 220 APV
1 APV ST12 100 APV.ST12
2 APV ST12 Elec 40 APV.ST12.Elec
3 APV ST12 Grocery 20 APV.ST12.Grocery
4 APV ST12 CPG 40 APV.ST12.CPG
</code></pre>
I am new to pandas and have spent lot of time but could not wrap my head around.

Concatenate values of columns in pandas

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有以下数据框：    Zone    Store   Department  TTLSales    0   APV                         220 1   APV     ST12                100 2   APV     ST12    Elec        40  ...

问连接pandas中列的值
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问连接pandas中列的值EN