bar*_*ons 3 python dataframe pandas
我有这样的数据框:
A B C D E
0 2 3 4 8 7
1 4 7 5 9 4
2 3 4 5 7 2
3 8 9 1 3 7
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我需要做这样的事情:
if 'value in column A' == 2:
'value for this row in new column' = 'value from column B' + 'value from column C'
elif 'value in column A' == 4:
'value for this row in new column' = 'value from column B' + 'value from column D'
elif 'value in column A' == 8:
'value for this row in new column' = 'value from column B' + 'value from column E'
else:
'value for this row in new column' = 0
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我尝试以几种方式做到这一点,例如:
1.
df['sum'][df['A'] == 2] = df['B'] + df['C']
df['sum'][df['A'] == 4] = df['B'] + df['D']
df['sum'][df['A'] == 8] = df['B'] + df['E']
2.
df.loc[df['A'] == 2, 'sum'] = df['B'] + df['C']
df.loc[df['A'] == 4, 'sum'] = df['B'] + df['D']
df.loc[df['A'] == 8, 'sum'] = df['B'] + df['E']
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但结果我有空单元格.
另一种简单的方法是使用字典并lookup获得总和即
colons = {2: 'C', 4: 'D', 8: 'E'}
df['sum']= np.nan
df['sum'] = df['B']+ df.lookup(df['A'].index,df['A'].map(colons).fillna('sum'))
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输出:
A B C D E sum 0 2 3 4 8 7 7.0 1 4 7 5 9 4 16.0 2 3 4 5 7 2 NaN 3 8 9 1 3 7 16.0
您可以使用0填充nan df.fillna(0)
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