Python 将多个空列添加到 Pandas DataFrame
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Add multiple empty columns to pandas DataFrame
提问by Winterflags
This may be a stupid question, but how do I add multiple empty columns to a DataFrame from a list?
这可能是一个愚蠢的问题，但是如何从列表向 DataFrame 添加多个空列？
I can do:
我可以：
df["B"] = None
df["C"] = None
df["D"] = None
But I can't do:
但我不能这样做：
df[["B", "C", "D"]] = None
KeyError: "['B' 'C' 'D'] not in index"
采纳答案by EdChum
I'd concatusing a DataFrame:
我会concat使用数据帧：
In [23]:
df = pd.DataFrame(columns=['A'])
df
Out[23]:
Empty DataFrame
Columns: [A]
Index: []
In [24]:
pd.concat([df,pd.DataFrame(columns=list('BCD'))])
Out[24]:
Empty DataFrame
Columns: [A, B, C, D]
Index: []
So by passing a list containing your original df, and a new one with the columns you wish to add, this will return a new df with the additional columns.
因此，通过传递一个包含原始 df 的列表和一个包含您希望添加的列的新列表，这将返回一个带有附加列的新 df。
Caveat: See the discussion of performance in the other answersand/or the comment discussions. reindexmay be preferable where performance is critical.
警告：请参阅其他答案和/或评论讨论中的性能讨论。reindex在性能至关重要的情况下可能更可取。
回答by unutbu
You could use df.reindexto add new columns:
您可以使用df.reindex添加新列：
In [18]: df = pd.DataFrame(np.random.randint(10, size=(5,1)), columns=['A'])
In [19]: df
Out[19]:
A
0 4
1 7
2 0
3 7
4 6
In [20]: df.reindex(columns=list('ABCD'))
Out[20]:
A B C D
0 4 NaN NaN NaN
1 7 NaN NaN NaN
2 0 NaN NaN NaN
3 7 NaN NaN NaN
4 6 NaN NaN NaN
reindexwill return a new DataFrame, with columns appearing in the order they are listed:
reindex将返回一个新的 DataFrame，列按它们列出的顺序出现：
In [31]: df.reindex(columns=list('DCBA'))
Out[31]:
D C B A
0 NaN NaN NaN 4
1 NaN NaN NaN 7
2 NaN NaN NaN 0
3 NaN NaN NaN 7
4 NaN NaN NaN 6
The reindexmethod as a fill_valueparameter as well:
该reindex方法也作为fill_value参数：
In [22]: df.reindex(columns=list('ABCD'), fill_value=0)
Out[22]:
A B C D
0 4 0 0 0
1 7 0 0 0
2 0 0 0 0
3 7 0 0 0
4 6 0 0 0
回答by toto_tico
If you don't want to rewrite the name of the old columns, then you can use reindex:
如果不想重写旧列的名称，则可以使用 reindex：
df.reindex(columns=[*df.columns.tolist(), 'new_column1', 'new_column2'], fill_value=0)
Full example:
完整示例：
In [1]: df = pd.DataFrame(np.random.randint(10, size=(3,1)), columns=['A'])
In [1]: df
Out[1]:
A
0 4
1 7
2 0
In [2]: df.reindex(columns=[*df.columns.tolist(), 'col1', 'col2'], fill_value=0)
Out[2]:
A col1 col2
0 1 0 0
1 2 0 0
And, if you already have a listwith the column names, :
而且，如果您已经有一个包含列名的列表，则：
In [3]: my_cols_list=['col1','col2']
In [4]: df.reindex(columns=[*df.columns.tolist(), *my_cols_list], fill_value=0)
Out[4]:
A col1 col2
0 1 0 0
1 2 0 0
回答by alexprice
Why not just use loop:
为什么不直接使用循环：
for newcol in ['B','C','D']:
df[newcol]=0
回答by Oleg O
Just to add to the list of funny ways:
只是添加到有趣的方式列表中：
columns_add = ['a', 'b', 'c']
df = df.assign(**dict(zip(columns_add, [0] * len(columns_add)))
