如何修改函数中的 Pandas DataFrame 以便调用者可以看到更改？
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How to modify a pandas DataFrame in a function so that changes are seen by the caller?
提问by ChaimG
I find myself doing repetitive tasks to various [pandas][1]DataFrames, so I made a function to do the processing. How do I modify dfin the function process_df(df)so that the caller sees all changes (without assigning a return value)?
我发现自己对各种[pandas][1]DataFrame执行重复性任务，因此我创建了一个函数来进行处理。如何df在函数中进行修改，process_df(df)以便调用者看到所有更改（不分配返回值）？
A simplified version of the code:
代码的简化版本：
def process_df(df):
df.columns = map(str.lower, df.columns)
df = pd.DataFrame({'A': [1], 'B': [2]})
process_df(df)
print df
A B 0 1 2
A B 0 1 2
EDIT new code:
编辑新代码：
def process_df(df):
df = df.loc[:, 'A']
df = pd.DataFrame({'A': [1], 'B': [2]})
process_df(df)
print df
A B 0 1 2
A B 0 1 2
采纳答案by Igor Raush
Indexing a DataFrameusing ix, loc, iloc, etc. returns a view of the underlying data (it is a read operation). In order to modify the contents of the frame you will need to use in-place transforms. For example,
DataFrame使用ix、loc、iloc等索引 a 会返回底层数据的视图（这是一个读取操作）。为了修改框架的内容，您需要使用就地变换。例如，
def process_df(df):
# drop all columns except for A
df.drop(df.columns[df.columns != 'A'], axis=1, inplace=True)
df = DataFrame({'A':[1,2,3], 'B':[1,2,3]})
process_df(df)
To change the order of columns, you can do something like this:
要更改列的顺序，您可以执行以下操作：
def process_df(df):
# swap A and B
df.columns = ['B', 'A']
df[['B', 'A']] = df[['A', 'B']]
