[英]python pandas how to multiplication columns by other values in different dataframe
[英]Python: How to replace only 0 values in a column by multiplication of 2 columns in Dataframe with a loop?
这是我的 Dataframe:
df={'pack':[2,2,2,2], 'a_cost':[10.5,0,11,0], 'b_cost':[0,6,0,6.5]}
它应该如下所示:
此时您会发现a_cost和b_cost列有 0,而其他列有值。 我希望我的 function 遵循这个逻辑......
for i in df.a_cost:
if i==0:
b_cost(column):value *(multiply) pack(column):value
replace 0 with this new multiplied value (example: 6.0*2=12)
for i in df_b.cost:
if i==0:
a_cost(column):value /(divide) pack(column):value
replace 0 with this new divided value (example: 10.5/2=5.25)
我无法弄清楚如何成功编写此逻辑......这是预期的output:
Output 代码:
df={'pack':[2,2,2,2], 'a_cost':[10.5,12.0,11,13.0], 'b_cost':[5.25,6,5.50,6.5]}
非常感谢您的帮助!
国际大学联合会,
df.loc[df.a_cost.eq(0), 'a_cost'] = df.b_cost * df.pack
df.loc[df.b_cost.eq(0), 'b_cost'] = df.a_cost / df.pack
您还可以使用mask
和fillna
:
df['a_cost'] = df.a_cost.mask(df.a_cost.eq(0)).fillna(df.b_cost * df.pack)
df['b_cost'] = df.b_cost.mask(df.b_cost.eq(0)).fillna(df.a_cost / df.pack)
按照评论更新,您可以在mask
中使用other
:
df['a_cost'] = df.a_cost.mask(df.a_cost.eq(0), other=df.b_cost * df.pack)
另请注意,一旦您已经在a_cost
列中填写了0
,则不需要第二次过滤。 也就是说,我们可以这样做:
df['b_cost'] = df.a_cost / df.pack
在这两种方法的第一个命令之后。
Output:
pack a_cost b_cost
0 2 10.5 5.25
1 2 12.0 6.00
2 2 11.0 5.50
3 2 13.0 6.50
尝试这个:
df['a_pack'] = df.apply(lambda x: x['b_cost']*x['pack'] if x['a_cost'] == 0 and x['b_cost'] != 0 else x['a_cost'], axis = 1)
df['b_pack'] = df.apply(lambda x: x['a_cost']/x['pack'] if x['b_cost'] == 0 and x['a_cost'] != 0 else x['b_cost'], axis = 1)
import numpy as np
df = pd.DataFrame({'pack':[2,2,2,2], 'a_cost':[10.5,0,11,0], 'b_cost':[0,6,0,6.5]})
df['a_cost'] = np.where(df['a_cost']==0, df['pack']*df['b_cost'], df['a_cost'])
df['b_cost'] = np.where(df['b_cost']==0, df['a_cost']/df['pack'], df['b_cost'])
print (df)
#pack a_cost b_cost
#0 2 10.5 5.25
#1 2 12.0 6.0
#2 2 11.0 5.50
#3 2 13.0 6.5
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