Python Pandas 将一列列表拆分为多列
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Pandas split column of lists into multiple columns
提问by user2938093
I have a pandas DataFrame with one column:
我有一个包含一列的 Pandas DataFrame：
import pandas as pd
df = pd.DataFrame(
data={
"teams": [
["SF", "NYG"],
["SF", "NYG"],
["SF", "NYG"],
["SF", "NYG"],
["SF", "NYG"],
["SF", "NYG"],
["SF", "NYG"],
]
}
)
print(df)
Output:
输出：
teams
0 [SF, NYG]
1 [SF, NYG]
2 [SF, NYG]
3 [SF, NYG]
4 [SF, NYG]
5 [SF, NYG]
6 [SF, NYG]
How can split this column of lists into 2 columns?
如何将此列列表拆分为 2 列？
采纳答案by jezrael
You can use DataFrameconstructor with listscreated by to_list:
您可以将DataFrame构造函数与listscreated by一起使用to_list：
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]
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
And for new DataFrame:
而对于新的DataFrame：
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
Solution with apply(pd.Series)is very slow:
解决方案apply(pd.Series)很慢：
#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)
回答by mikkokotila
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:
与建议的解决方案相反，似乎有一种语法更简单的方法，因此更容易记住。我假设该列在数据帧 df 中称为“元”：
df2 = pd.DataFrame(df['meta'].str.split().values.tolist())
回答by Joseph Davison
Much simpler solution:
更简单的解决方案：
pd.DataFrame(df2["teams"].to_list(), columns=['team1', 'team2'])
Yields,
产量，
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
If you wanted to split a column of delimited strings rather than lists, you could similarly do:
如果您想拆分一列分隔字符串而不是列表，您可以类似地执行以下操作：
pd.DataFrame(df["teams"].str.split('<delim>', expand=True).values,
columns=['team1', 'team2'])
回答by Kevin Markham
This solution preserves the index of the df2DataFrame, unlike any solution that uses tolist():
此解决方案保留了df2DataFrame的索引，与使用tolist()以下任何解决方案不同：
df3 = df2.teams.apply(pd.Series)
df3.columns = ['team1', 'team2']
Here's the result:
结果如下：
team1 team2
0 SF NYG
1 SF NYG
2 SF NYG
3 SF NYG
4 SF NYG
5 SF NYG
6 SF NYG
回答by ailurid
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:
基于之前的答案，这里是另一个解决方案，它返回与 df2.teams.apply(pd.Series) 相同的结果，运行时间要快得多：
pd.DataFrame([{x: y for x, y in enumerate(item)} for item in df2['teams'].values.tolist()], index=df2.index)
Timings:
时间：
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)
回答by Lucas
The above solutions didn't work for me since I have nanobservations in my dataframe. In my case df2[['team1','team2']] = pd.DataFrame(df2.teams.values.tolist(), index= df2.index)yields:
上述解决方案对我不起作用，因为我nan在我的dataframe. 在我的情况下df2[['team1','team2']] = pd.DataFrame(df2.teams.values.tolist(), index= df2.index)产生：
object of type 'float' has no len()
I solve this using list comprehension. Here the replicable example:
我使用列表理解来解决这个问题。这是可复制的示例：
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
output:
输出：
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
solving with list comprehension:
用列表理解解决：
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
yields:
产量：
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
回答by Talis
list comprehension
列表理解
simple implementation with list comprehension ( my favorite)
列表理解的简单实现（我最喜欢的）
df = pd.DataFrame([pd.Series(x) for x in df.teams])
df.columns = ['team_{}'.format(x+1) for x in df.columns]
timing on output:
输出时间：
CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 2.71 ms
output:
输出：
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
