pandas 如何在合并熊猫数据框中的两列时删除 nan 值？
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How to remove nan value while combining two column in Panda Data frame?
提问by imSonuGupta
I am trying but not able to remove nanwhile combining two columns of a DataFrame.
我正在尝试但无法nan在组合 .a 的两列时删除DataFrame。
Data is like:
数据是这样的：
feedback_id _id
568a8c25cac4991645c287ac nan
568df45b177e30c6487d3603 nan
nan 568df434832b090048f34974
nan 568cd22e9e82dfc166d7dff1
568df3f0832b090048f34711 nan
nan 568e5a38b4a797c664143dda
I want:
我想要：
feedback_request_id
568a8c25cac4991645c287ac
568df45b177e30c6487d3603
568df434832b090048f34974
568cd22e9e82dfc166d7dff1
568df3f0832b090048f34711
568e5a38b4a797c664143dda
Here is my code:
这是我的代码：
df3['feedback_request_id'] = ('' if df3['_id'].empty else df3['_id'].map(str)) + ('' if df3['feedback_id'].empty else df3['feedback_id'].map(str))
Output I'm getting:
我得到的输出：
feedback_request_id
568a8c25cac4991645c287acnan
568df45b177e30c6487d3603nan
nan568df434832b090048f34974
nan568cd22e9e82dfc166d7dff1
568df3f0832b090048f34711nan
nan568e5a38b4a797c664143dda
I have tried this, also:
我也试过这个：
df3['feedback_request_id'] = ('' if df3['_id']=='nan' else df3['_id'].map(str)) + ('' if df3['feedback_id']=='nan' else df3['feedback_id'].map(str))
But it's giving the error:
但它给出了错误：
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
回答by jezrael
You can use combine_firstor fillna:
您可以使用combine_first或fillna：
print df['feedback_id'].combine_first(df['_id'])
0 568a8c25cac4991645c287ac
1 568df45b177e30c6487d3603
2 568df434832b090048f34974
3 568cd22e9e82dfc166d7dff1
4 568df3f0832b090048f34711
5 568e5a38b4a797c664143dda
Name: feedback_id, dtype: object
print df['feedback_id'].fillna(df['_id'])
0 568a8c25cac4991645c287ac
1 568df45b177e30c6487d3603
2 568df434832b090048f34974
3 568cd22e9e82dfc166d7dff1
4 568df3f0832b090048f34711
5 568e5a38b4a797c664143dda
Name: feedback_id, dtype: object
回答by BallpointBen
If you want a solution that doesn't require referencing dftwice or any of its columns explicitly:
如果您想要一个不需要df明确引用两次或其任何列的解决方案：
df.bfill(axis=1).iloc[:, 0]
With two columns, this will copy non-null values from the right column into the left, then select the left column.
对于两列，这会将非空值从右列复制到左列，然后选择左列。
回答by jpp
For an in-place solution, you can use pd.Series.updatewith pd.DataFrame.pop:
对于就地解决方案，您可以使用pd.Series.updatewith pd.DataFrame.pop：
df['feedback_id'].update(df.pop('_id'))
print(df)
feedback_id
0 568a8c25cac4991645c287ac
1 568df45b177e30c6487d3603
2 568df434832b090048f34974
3 568cd22e9e82dfc166d7dff1
4 568df3f0832b090048f34711
5 568e5a38b4a797c664143dda
