4

我有一个df包含'TPrice','THigh','TLow','TOpen','TClose','TPCLOSE'列的数据框，现在我想将'TPrice','THigh','TLow','TOpen','TClose'列值设置为与列值为零'TPCLOSE'的行的列相同。TPrice

显示一些 TPrice 为 0 的行：

>>> df[df['TPrice']==0][['TPrice','THigh','TLow','TOpen','TClose','TPCLOSE']][0:5]
    TPrice  THigh  TLow  TOpen  TClose  TPCLOSE
13       0      0     0      0       0     4.19
19       0      0     0      0       0     7.74
32       0      0     0      0       0     3.27
43       0      0     0      0       0    12.98
60       0      0     0      0       0     7.48

然后分配：

>>> df[df['TPrice']==0][['TPrice','THigh','TLow','TOpen','TClose']] = df['TPCLOSE']

但是 Pandas 并没有真正改变 df ，因为下面的代码仍然可以找到一些行：

>>> df[df['TPrice']==0][['TPrice','THigh','TLow','TOpen','TClose','TPCLOSE']][0:5]
    TPrice  THigh  TLow  TOpen  TClose  TPCLOSE
13       0      0     0      0       0     4.19
19       0      0     0      0       0     7.74
32       0      0     0      0       0     3.27
43       0      0     0      0       0    12.98
60       0      0     0      0       0     7.48

那么怎么办呢？

Jeff 解决方案的更新：

>>> quote_df = get_quote()
>>> quote_df[quote_df['TPrice']==0][['TPrice','THigh','TLow','TOpen','TClose','TPCLOSE','RT','TVol']][0:5]
    TPrice  THigh  TLow  TOpen  TClose  TPCLOSE   RT  TVol
13       0      0     0      0       0     4.19 -100     0
32       0      0     0      0       0     3.27 -100     0
43       0      0     0      0       0    12.98 -100     0
45       0      0     0      0       0    26.74 -100     0
60       0      0     0      0       0     7.48 -100     0
>>> row_selection = quote_df['TPrice']==0
>>> col_selection = ['THigh','TLow','TOpen','TClose']
>>> for col in col_selection:
...     quote_df.loc[row_selection, col] = quote_df['TPCLOSE']
... 
>>> quote_df[quote_df['TPrice']==0][['TPrice','THigh','TLow','TOpen','TClose','TPCLOSE','RT','TVol']][0:5]
    TPrice  THigh  TLow  TOpen  TClose  TPCLOSE   RT  TVol
13       0   4.19  4.19   4.19    4.19     4.19 -100     0
32       0   4.19  4.19   4.19    4.19     3.27 -100     0
43       0   4.19  4.19   4.19    4.19    12.98 -100     0
45       0   4.19  4.19   4.19    4.19    26.74 -100     0
60       0   4.19  4.19   4.19    4.19     7.48 -100     0
>>>
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2 回答 2

4

这个操作不会自动广播，所以你需要做这样的事情

In [17]: df = DataFrame(dict(A = [1,2,0,0,0],B=[0,0,0,10,11],C=[3,4,5,6,7]))

In [18]: df
Out[18]: 
   A   B  C
0  1   0  3
1  2   0  4
2  0   0  5
3  0  10  6
4  0  11  7

如果您正在修改 A（就像您在这里一样），请计算您要首先屏蔽哪些行（否则它们可能会随着您的变化而改变）

In [19]: mask = df['A'] == 0

In [20]: for col in ['A','B']:
   ....:     df.loc[mask,col] = df['C']
   ....:     

In [21]: df
Out[21]: 
   A  B  C
0  1  0  3
1  2  0  4
2  5  5  5
3  6  6  6
4  7  7  7

这需要进行更改以使其更自然（因为您正在将 rhs 上的系列分配给 lhs 上的数据帧，现在不会像您认为的那样广播） https://github.com/pydata/熊猫/问题/5206

于 2013-10-13T15:33:35.187 回答
1 
>>> import pandas as pd
>>> test=pd.DataFrame({'A': [0,1,2], 'B': [3,4,5], 'C': [6,7,8]})
>>> test
   A  B  C
0  0  3  6
1  1  4  7
2  2  5  8
>>> test.apply(lambda x: x.where(test.A!=0, test.C), axis=0)
   A  B  C
0  6  6  6
1  1  4  7
2  2  5  8
于 2013-10-13T14:30:16.600 回答