blocks|key|429381|text|您可以将DataFrame构造函数与to_list创建的lists一起使用。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|429382|import+pandas+as+pd

d1+=+{'teams':+[['SF',+'NYG'],['SF',+'NYG'],['SF',+'NYG'],
++++++++++++++++['SF',+'NYG'],['SF',+'NYG'],['SF',+'NYG'],['SF',+'NYG']]}
df2+=+pd.DataFrame(d1)
print+(df2)
+++++++teams
0++[SF,+NYG]
1++[SF,+NYG]
2++[SF,+NYG]
3++[SF,+NYG]
4++[SF,+NYG]
5++[SF,+NYG]
6++[SF,+NYG]|code-block|syntax|javascript|429383|df2[['team1','team2']]+=+pd.DataFrame(df2.teams.tolist(),+index=+df2.index)
print+(df2)
+++++++teams+team1+team2
0++[SF,+NYG]++++SF+++NYG
1++[SF,+NYG]++++SF+++NYG
2++[SF,+NYG]++++SF+++NYG
3++[SF,+NYG]++++SF+++NYG
4++[SF,+NYG]++++SF+++NYG
5++[SF,+NYG]++++SF+++NYG
6++[SF,+NYG]++++SF+++NYG|429384|为了一个新的DataFrame|429385|df3+=+pd.DataFrame(df2['teams'].to_list(),+columns=['team1','team2'])
print+(df3)
++team1+team2
0++++SF+++NYG
1++++SF+++NYG
2++++SF+++NYG
3++++SF+++NYG
4++++SF+++NYG
5++++SF+++NYG
6++++SF+++NYG|429386|使用apply(pd.Series)的解决方案非常慢：|429387|#7k+rows
df2+=+pd.concat([df2]*1000).reset_index(drop=True)

In+[121]:+%25timeit+df2['teams'].apply(pd.Series)
1.79+s+±+52.5+ms+per+loop+(mean+±+std.+dev.+of+7+runs,+1+loop+each)

In+[122]:+%25timeit+pd.DataFrame(df2['teams'].to_list(),+columns=['team1','team2'])
1.63+ms+±+54.3+µs+per+loop+(mean+±+std.+dev.+of+7+runs,+1000+loops+each)|429388|entityMap|0|LINK|mutability|MUTABLE|url|https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.to_list.html^0|4|9|I|7|S|5|I|7|0|0|0|0|6|9|0|0|2|G|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@$9|19|A|1A|1|1B]]|E|$]]|$1|F|3|G|5|H|7|1C|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|H|7|1D|8|@]|D|@]|E|$I|J]]|$1|M|3|N|5|6|7|1E|8|@$9|1F|A|1G|B|C]]|D|@]|E|$]]|$1|O|3|P|5|H|7|1H|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|1I|8|@$9|1J|A|1K|B|C]]|D|@]|E|$]]|$1|S|3|T|5|H|7|1L|8|@]|D|@]|E|$I|J]]|$1|U|3|-4|5|6|7|1M|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11]]]]

You can use the <code>DataFrame</code> constructor with <code>lists</code> created by <a href="https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.to_list.html" rel="noreferrer"><code>to_list</code></a>:
<pre><code>import pandas as pd

d1 = {'teams': [['SF', 'NYG'],['SF', 'NYG'],['SF', 'NYG'],
 ['SF', 'NYG'],['SF', 'NYG'],['SF', 'NYG'],['SF', 'NYG']]}
df2 = pd.DataFrame(d1)
print (df2)
 teams
0 [SF, NYG]
1 [SF, NYG]
2 [SF, NYG]
3 [SF, NYG]
4 [SF, NYG]
5 [SF, NYG]
6 [SF, NYG]
</code></pre>
<hr />
<pre><code>df2[['team1','team2']] = pd.DataFrame(df2.teams.tolist(), index= df2.index)
print (df2)
 teams team1 team2
0 [SF, NYG] SF NYG
1 [SF, NYG] SF NYG
2 [SF, NYG] SF NYG
3 [SF, NYG] SF NYG
4 [SF, NYG] SF NYG
5 [SF, NYG] SF NYG
6 [SF, NYG] SF NYG
</code></pre>
And for a new <code>DataFrame</code>:
<pre><code>df3 = pd.DataFrame(df2['teams'].to_list(), columns=['team1','team2'])
print (df3)
 team1 team2
0 SF NYG
1 SF NYG
2 SF NYG
3 SF NYG
4 SF NYG
5 SF NYG
6 SF NYG
</code></pre>
A solution with <code>apply(pd.Series)</code> is very slow:
<pre><code>#7k rows
df2 = pd.concat([df2]*1000).reset_index(drop=True)

In [121]: %timeit df2['teams'].apply(pd.Series)
1.79 s ± 52.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [122]: %timeit pd.DataFrame(df2['teams'].to_list(), columns=['team1','team2'])
1.63 ms ± 54.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
</code></pre>

blocks|key|2457666|text|似乎有一种语法上更简单的方法，因此更容易记住，而不是建议的解决方案。我假设该列在dataframe中称为“meta”：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2457667|df2+=+pd.DataFrame(df['meta'].str.split().values.tolist())|code-block|syntax|javascript|2457668|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

There seems to be a syntactically simpler way, and therefore easier to remember, as opposed to the proposed solutions. I'm assuming that the column is called 'meta' in a dataframe df: 

<pre><code>df2 = pd.DataFrame(df['meta'].str.split().values.tolist())
</code></pre>

blocks|key|242782|text|更简单的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|242783|pd.DataFrame(df2["teams"].to_list(),+columns=['team1',+'team2'])|code-block|syntax|javascript|242784|收益，|242785|++team1+team2
-------------
0++++SF+++NYG
1++++SF+++NYG
2++++SF+++NYG
3++++SF+++NYG
4++++SF+++NYG
5++++SF+++NYG
6++++SF+++NYG
7++++SF+++NYG|242786|如果您想拆分分隔字符串的列而不是列表，也可以这样做：|242787|pd.DataFrame(df["teams"].str.split('<delim>',+expand=True).values,
+++++++++++++columns=['team1',+'team2'])|242788|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Much simpler solution:

<pre><code>pd.DataFrame(df2["teams"].to_list(), columns=['team1', 'team2'])
</code></pre>

Yields,

<pre><code> team1 team2
-------------
0 SF NYG
1 SF NYG
2 SF NYG
3 SF NYG
4 SF NYG
5 SF NYG
6 SF NYG
7 SF NYG
</code></pre>

If you wanted to split a column of delimited strings rather than lists, you could similarly do:

<pre><code>pd.DataFrame(df["teams"].str.split('&lt;delim&gt;', expand=True).values,
 columns=['team1', 'team2'])
</code></pre>

blocks|key|429502|text|此解决方案保留df2+DataFrame的索引，不像任何使用tolist()的解决方案。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|429503|df3+=+df2.teams.apply(pd.Series)
df3.columns+=+['team1',+'team2']|code-block|syntax|javascript|429504|结果如下：|429505|++team1+team2
0++++SF+++NYG
1++++SF+++NYG
2++++SF+++NYG
3++++SF+++NYG
4++++SF+++NYG
5++++SF+++NYG
6++++SF+++NYG|429506|entityMap^0|7|3|U|8|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

This solution preserves the index of the <code>df2</code> DataFrame, unlike any solution that uses <code>tolist()</code>:

<pre class="lang-py prettyprint-override"><code>df3 = df2.teams.apply(pd.Series)
df3.columns = ['team1', 'team2']
</code></pre>

Here's the result:

<pre><code> team1 team2
0 SF NYG
1 SF NYG
2 SF NYG
3 SF NYG
4 SF NYG
5 SF NYG
6 SF NYG
</code></pre>

blocks|key|2457725|text|基于前面的答案，下面是另一个解决方案，它返回与df2.teams.apply(pd.Series)相同的运行时间更快的结果：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2457726|pd.DataFrame([{x:+y+for+x,+y+in+enumerate(item)}+for+item+in+df2['teams'].values.tolist()],+index=df2.index)|code-block|syntax|javascript|2457727|计时：|2457728|In+[1]:
import+pandas+as+pd
d1+=+{'teams':+[['SF',+'NYG'],['SF',+'NYG'],['SF',+'NYG'],
++++++++++++++++['SF',+'NYG'],['SF',+'NYG'],['SF',+'NYG'],['SF',+'NYG']]}
df2+=+pd.DataFrame(d1)
df2+=+pd.concat([df2]*1000).reset_index(drop=True)

In+[2]:+%25timeit+df2['teams'].apply(pd.Series)

8.27+s+±+2.73+s+per+loop+(mean+±+std.+dev.+of+7+runs,+1+loop+each)

In+[3]:+%25timeit+pd.DataFrame([{x:+y+for+x,+y+in+enumerate(item)}+for+item+in+df2['teams'].values.tolist()],+index=df2.index)

35.4+ms+±+5.22+ms+per+loop+(mean+±+std.+dev.+of+7+runs,+10+loops+each)|2457729|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|$]]

Based on the previous answers, here is another solution which returns the same result as df2.teams.apply(pd.Series) with a much faster run time:

<pre><code>pd.DataFrame([{x: y for x, y in enumerate(item)} for item in df2['teams'].values.tolist()], index=df2.index)
</code></pre>

Timings:

<pre><code>In [1]:
import pandas as pd
d1 = {'teams': [['SF', 'NYG'],['SF', 'NYG'],['SF', 'NYG'],
 ['SF', 'NYG'],['SF', 'NYG'],['SF', 'NYG'],['SF', 'NYG']]}
df2 = pd.DataFrame(d1)
df2 = pd.concat([df2]*1000).reset_index(drop=True)

In [2]: %timeit df2['teams'].apply(pd.Series)

8.27 s ± 2.73 s per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [3]: %timeit pd.DataFrame([{x: y for x, y in enumerate(item)} for item in df2['teams'].values.tolist()], index=df2.index)

35.4 ms ± 5.22 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
</code></pre>

blocks|key|429520|text|以前的解决方案对我没有用，因为我的dataframe中有dataframe观察结果。在我的例子中，df2[['team1','team2']]+=+pd.DataFrame(df2.teams.values.tolist(),+index=+df2.index)产生了：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|429521|object+of+type+'float'+has+no+len()|code-block|syntax|javascript|429522|我用一个列表理解来解决这个问题。以下是可复制的例子：|429523|import+pandas+as+pd
import+numpy+as+np
d1+=+{'teams':+[['SF',+'NYG'],['SF',+'NYG'],['SF',+'NYG'],
++++++++++++['SF',+'NYG'],['SF',+'NYG'],['SF',+'NYG'],['SF',+'NYG']]}
df2+=+pd.DataFrame(d1)
df2.loc[2,'teams']+=+np.nan
df2.loc[4,'teams']+=+np.nan
df2|429524|输出：|429525|++++++++teams
0+++[SF,+NYG]
1+++[SF,+NYG]
2+++NaN
3+++[SF,+NYG]
4+++NaN
5+++[SF,+NYG]
6+++[SF,+NYG]

df2['team1']=np.nan
df2['team2']=np.nan|429526|用清单理解来解决问题，|429527|for+i+in+[0,1]:
++++df2['team{}'.format(str(i%2B1))]=[k[i]+if+isinstance(k,list)+else+k+for+k+in+df2['teams']]

df2|429528|产量：|429529|++++teams+++team1+++team2
0+++[SF,+NYG]+++SF++NYG
1+++[SF,+NYG]+++SF++NYG
2+++NaN++++++++NaN++NaN
3+++[SF,+NYG]+++SF++NYG
4+++NaN++++++++NaN++NaN
5+++[SF,+NYG]+++SF++NYG
6+++[SF,+NYG]+++SF++NYG|429530|entityMap^0|H|9|S|9|1D|2A|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|19|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1A|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1C|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1D|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|1E|8|@]|D|@]|E|$]]|$1|U|3|V|5|H|7|1F|8|@]|D|@]|E|$I|J]]|$1|W|3|X|5|6|7|1G|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|H|7|1H|8|@]|D|@]|E|$I|J]]|$1|10|3|-4|5|6|7|1I|8|@]|D|@]|E|$]]]|11|$]]

The previous solutions didn't work for me since I have <code>nan</code> observations in my <code>dataframe</code>. In my case <code>df2[['team1','team2']] = pd.DataFrame(df2.teams.values.tolist(), index= df2.index)</code> yields:
<pre><code>object of type 'float' has no len()
</code></pre>
I solve this using a list comprehension. Here is the replicable example:
<pre><code>import pandas as pd
import numpy as np
d1 = {'teams': [['SF', 'NYG'],['SF', 'NYG'],['SF', 'NYG'],
 ['SF', 'NYG'],['SF', 'NYG'],['SF', 'NYG'],['SF', 'NYG']]}
df2 = pd.DataFrame(d1)
df2.loc[2,'teams'] = np.nan
df2.loc[4,'teams'] = np.nan
df2
</code></pre>
Output:
<pre><code> teams
0 [SF, NYG]
1 [SF, NYG]
2 NaN
3 [SF, NYG]
4 NaN
5 [SF, NYG]
6 [SF, NYG]

df2['team1']=np.nan
df2['team2']=np.nan
</code></pre>
Solving with a list comprehension,
<pre><code>for i in [0,1]:
 df2['team{}'.format(str(i+1))]=[k[i] if isinstance(k,list) else k for k in df2['teams']]

df2
</code></pre>
yields:
<pre><code> teams team1 team2
0 [SF, NYG] SF NYG
1 [SF, NYG] SF NYG
2 NaN NaN NaN
3 [SF, NYG] SF NYG
4 NaN NaN NaN
5 [SF, NYG] SF NYG
6 [SF, NYG] SF NYG
</code></pre>

blocks|key|2457765|text|列表理解|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|2457766|一个简单的列表理解实现(我最喜欢)|2457767|df+=+pd.DataFrame([pd.Series(x)+for+x+in+df.teams])
df.columns+=+['team_{}'.format(x%2B1)+for+x+in+df.columns]|code-block|syntax|javascript|2457768|产出的时间安排：|2457769|CPU+times:+user+0+ns,+sys:+0+ns,+total:+0+ns
Wall+time:+2.71+ms|2457770|输出：|2457771|team_1++++team_2
0++++SF++++NYG
1++++SF++++NYG
2++++SF++++NYG
3++++SF++++NYG
4++++SF++++NYG
5++++SF++++NYG
6++++SF++++NYG|2457772|entityMap^0|0|4|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|11|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|13|8|@]|D|@]|E|$]]|$1|S|3|T|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|U|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|V|$]]

List comprehension
A simple implementation with list comprehension (my favorite)
<pre><code>df = pd.DataFrame([pd.Series(x) for x in df.teams])
df.columns = ['team_{}'.format(x+1) for x in df.columns]
</code></pre>
Timing on output:
<pre><code>CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 2.71 ms

</code></pre>
Output:
<pre><code>team_1 team_2
0 SF NYG
1 SF NYG
2 SF NYG
3 SF NYG
4 SF NYG
5 SF NYG
6 SF NYG
</code></pre>

blocks|key|429600|text|下面是使用df.transform和df.set_index的另一种解决方案|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|429601|>>>+from+operator+import+itemgetter
>>>+df['teams'].transform({'item1':+itemgetter(0),+'item2':+itemgetter(1)})

++team1+team2
0++++SF+++NYG
1++++SF+++NYG
2++++SF+++NYG
3++++SF+++NYG
4++++SF+++NYG
5++++SF+++NYG
6++++SF+++NYG|code-block|syntax|javascript|429602|这当然可以概括为：|429603|>>>+indices+=+range(len(df['teams'][0]))

>>>+df['teams'].transform({f'team{i%2B1}':+itemgetter(i)+for+i+in+indices})

++team1+team2
0++++SF+++NYG
1++++SF+++NYG
2++++SF+++NYG
3++++SF+++NYG
4++++SF+++NYG
5++++SF+++NYG
6++++SF+++NYG|429604|这种方法的另一个好处是提取所需的指数：|429605|>>>+df
+++++++++++++++++teams
0++[SF,+NYG,+XYZ,+ABC]
1++[SF,+NYG,+XYZ,+ABC]
2++[SF,+NYG,+XYZ,+ABC]
3++[SF,+NYG,+XYZ,+ABC]
4++[SF,+NYG,+XYZ,+ABC]
5++[SF,+NYG,+XYZ,+ABC]
6++[SF,+NYG,+XYZ,+ABC]

>>>+indices+=+[0,+2]
>>>+df['teams'].transform({f'team{i%2B1}':+itemgetter(i)+for+i+in+indices})

++team1+team3
0++++SF+++XYZ
1++++SF+++XYZ
2++++SF+++XYZ
3++++SF+++XYZ
4++++SF+++XYZ
5++++SF+++XYZ
6++++SF+++XYZ|429606|entityMap|0|LINK|mutability|MUTABLE|url|https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.transform.html|1|https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.set_index.html^0|5|C|I|C|5|C|0|I|C|1|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@$9|17|A|18|1|19]|$9|1A|A|1B|1|1C]]|E|$]]|$1|F|3|G|5|H|7|1D|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1E|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1F|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1G|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1H|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|1I|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z]]|10|$5|V|W|X|E|$Y|11]]]]

Here's another solution using <a href="https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.transform.html" rel="nofollow noreferrer"><code>df.transform</code></a> and <a href="https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.set_index.html" rel="nofollow noreferrer"><code>df.set_index</code></a>:
<pre><code>&gt;&gt;&gt; from operator import itemgetter
&gt;&gt;&gt; df['teams'].transform({'item1': itemgetter(0), 'item2': itemgetter(1)})

 team1 team2
0 SF NYG
1 SF NYG
2 SF NYG
3 SF NYG
4 SF NYG
5 SF NYG
6 SF NYG
</code></pre>
Which of course can be generalized as:
<pre><code>&gt;&gt;&gt; indices = range(len(df['teams'][0]))

&gt;&gt;&gt; df['teams'].transform({f'team{i+1}': itemgetter(i) for i in indices})

 team1 team2
0 SF NYG
1 SF NYG
2 SF NYG
3 SF NYG
4 SF NYG
5 SF NYG
6 SF NYG
</code></pre>
This approach has the added benefit of extracting the desired indices:
<pre><code>&gt;&gt;&gt; df
 teams
0 [SF, NYG, XYZ, ABC]
1 [SF, NYG, XYZ, ABC]
2 [SF, NYG, XYZ, ABC]
3 [SF, NYG, XYZ, ABC]
4 [SF, NYG, XYZ, ABC]
5 [SF, NYG, XYZ, ABC]
6 [SF, NYG, XYZ, ABC]

&gt;&gt;&gt; indices = [0, 2]
&gt;&gt;&gt; df['teams'].transform({f'team{i+1}': itemgetter(i) for i in indices})

 team1 team3
0 SF XYZ
1 SF XYZ
2 SF XYZ
3 SF XYZ
4 SF XYZ
5 SF XYZ
6 SF XYZ
</code></pre>

blocks|key|429627|text|我想推荐一种更有效的和毕达通的方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|429628|首先，将DataFrame定义为原始帖子：|429629|df+=+pd.DataFrame({"teams":+[["SF",+"NYG"]+for+_+in+range(7)]})|code-block|syntax|javascript|429630|我的解决方案：|429631|%25%25timeit
df['team1'],+df['team2']+=+zip(*list(df['teams'].values))
>>+761+µs+±+8.35+µs+per+loop|429632|相比之下，最受欢迎的解决方案是：|429633|%25%25timeit
df[['team1','team2']]+=+pd.DataFrame(df.teams.tolist(),+index=df.index)
df+=+pd.DataFrame(df['teams'].to_list(),+columns=['team1','team2'])
>>+1.31+ms+±+11.2+µs+per+loop|429634|我的解决方案节省了40%25的时间和更短的时间。唯一需要记住的是如何使用zip(*list)解压缩和重塑二维列表。|offset|length|style|CODE|429635|entityMap^0|0|0|0|0|0|0|0|Y|A|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|11|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|13|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|15|8|@$S|16|T|17|U|V]]|9|@]|A|$]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

I would like to recommend a more efficient and Pythonic way.
First define the DataFrame as original post:
<pre class="lang-py prettyprint-override"><code>df = pd.DataFrame({&quot;teams&quot;: [[&quot;SF&quot;, &quot;NYG&quot;] for _ in range(7)]})
</code></pre>
My solution:
<pre class="lang-py prettyprint-override"><code>%%timeit
df['team1'], df['team2'] = zip(*list(df['teams'].values))
&gt;&gt; 761 µs ± 8.35 µs per loop
</code></pre>
In comparison, the most upvoted solution:
<pre class="lang-py prettyprint-override"><code>%%timeit
df[['team1','team2']] = pd.DataFrame(df.teams.tolist(), index=df.index)
df = pd.DataFrame(df['teams'].to_list(), columns=['team1','team2'])
&gt;&gt; 1.31 ms ± 11.2 µs per loop
</code></pre>
My solution saves 40% time and is much shorter. The only thing you need to remember is how to unpack and reshape a two-dimension list by using <code>zip(*list)</code>.

blocks|key|429668|text|您可以尝试使用两次应用程序在df中创建新列“team1”和“team2”。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|429669|df+=+pd.DataFrame({"teams":+[["SF",+"NYG"]+for+_+in+range(7)]})
df["team1"]=df['teams'].apply(lambda+x:+x[0]++)
df["team2"]=df['teams'].apply(lambda+x:+x[1]++)
df|code-block|syntax|javascript|429670|在这里输入图像描述|offset|length|429671|entityMap|0|LINK|mutability|MUTABLE|url|https://i.stack.imgur.com/4R05A.png^0|0|0|0|9|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

you can try to use two times of apply to create new column 'team1' and 'team2' in your df
<pre><code>df = pd.DataFrame({&quot;teams&quot;: [[&quot;SF&quot;, &quot;NYG&quot;] for _ in range(7)]})
df[&quot;team1&quot;]=df['teams'].apply(lambda x: x[0] )
df[&quot;team2&quot;]=df['teams'].apply(lambda x: x[1] )
df
</code></pre>
<a href="https://i.stack.imgur.com/4R05A.png" rel="nofollow noreferrer">enter image description here</a>

I have a Pandas DataFrame with one column:
<pre><code>import pandas as pd

df = pd.DataFrame({&quot;teams&quot;: [[&quot;SF&quot;, &quot;NYG&quot;] for _ in range(7)]})

 teams
0 [SF, NYG]
1 [SF, NYG]
2 [SF, NYG]
3 [SF, NYG]
4 [SF, NYG]
5 [SF, NYG]
6 [SF, NYG]
</code></pre>
How can split this column of lists into two columns?
Desired result:
<pre><code> team1 team2
0 SF NYG
1 SF NYG
2 SF NYG
3 SF NYG
4 SF NYG
5 SF NYG
6 SF NYG
</code></pre>

Split a Pandas column of lists into multiple columns

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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我有一个Pandas DataFrame，有一个专栏：import pandas as pddf = pd.DataFrame({"teams": [["SF", "NYG"] for _ in range(7)]})       teams0  [SF, NYG]1  [SF, NYG]2  [SF, NYG]3  ...

问将Pandas列表列拆分为多个列
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问将Pandas列表列拆分为多个列EN