如何从数字中减去 Pandas DataFrame 的每一行？
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How to subtract each row of a pandas DataFrame from a number?
提问by dana111
Say I have a DataFrame:
假设我有一个 DataFrame：
ds = pd.DataFrame(np.abs(randn(3, 4)), index=[1,2,3], columns=['A','B','C','Average'])
ds
A B C Average
1 1.099679 0.042043 0.083903 0.410128
2 0.268205 0.718933 1.459374 0.758887
3 0.680566 0.538655 0.038236 1.169403
How do I subtract (and replace with the result) A, B and C in row one with the average in row 1?
如何减去（并用结果替换）第一行中的 A、B 和 C 与第一行中的平均值？
回答by Alex Riley
One relatively simple way is to use the submethod (I'm assuming that Averageis always the last column):
一种相对简单的方法是使用该sub方法（我假设它Average始终是最后一列）：
ds[ds.columns[:-1]].sub(ds.Average, axis=0)
This does the following:
这将执行以下操作：
ds[ds.columns[:-1]]is a DataFrame containing all but the last column (Average) ofds..sub(ds.Average, axis=0)subtracts the row-values in theAveragecolumn from the corresponding rows in the DataFrame.
ds[ds.columns[:-1]]是一个包含除最后一列 (Average)之外的所有数据帧ds。.sub(ds.Average, axis=0)从AverageDataFrame 中的相应行中减去列中的行值。
To alter your original ds, make sure to rebind the relevant columns of dsto the new DataFrame of values:
要更改您的原始ds，请确保将 的相关列重新绑定ds到新的 DataFrame 值：
ds[ds.columns[:-1]] = ds[ds.columns[:-1]].sub(ds.Average, axis=0)
回答by Woody Pride
How about
怎么样
ds['A'] = ds['A'] - ds['Average']
ds['B'] = ds['B'] - ds['Average']
ds['C'] = ds['C'] - ds['Average']
Pandas is easy like that!
Pandas就是这么简单！
Oh, that does it for the entire DF. You only want it for the firs row is that right?
哦，这对整个 DF 都是如此。你只想要第一行是吗？
ds.loc[1, 'A'] = ds.loc[1, 'A'] - ds.loc[1, 'Average']
ds.loc[1, 'B'] = ds.loc[1, 'B'] - ds.loc[1, 'Average']
ds.loc[1, 'C'] = ds.loc[1, 'C'] - ds.loc[1, 'Average']
or in a loop:
或在循环中：
for col in ['A', 'B', 'C']:
ds.loc[1, col] = df.loc[1, col] - ds.loc[1, 'Average']
and so on...
等等...
if you have thousands of columns then simply do:
如果您有数千列，那么只需执行以下操作：
for col in ds.columns:
ds[col] = ds[col] - ds['Average']
