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Pandas-将列除以另一列,条件是值是否大于0?

[英]Pandas- Dividing a column by another column conditional on if values are greater than 0?

I have a pandas dataframe that contains dates, items, and 2 values. 我有一个包含日期,项目和2个值的pandas数据框。 All I'm looking to do is output another column that is the product of column A / column B if column B is greater than 0, and 0 if column B is equal to 0. 我要做的就是输出另一列,如果列B大于0则是列A /列B的乘积,如果列B等于0,则输出0。

   date     item   A   B        C       
 1/1/2017   a      0   3             0  
 1/1/2017   b      2   0             0  
 1/1/2017   c      5   2           2.5  
 1/1/2017   d      4   1             4  
 1/1/2017   e      3   3             1  
 1/1/2017   f      0   4             0  
 1/2/2017   a      3   3             1  
 1/2/2017   b      2   2             1  
 1/2/2017   c      3   9   0.333333333  
 1/2/2017   d      4   0             0  
 1/2/2017   e      5   3   1.666666667  
 1/2/2017   f      3   0             0  

this is the code I've written, but the kernel keeps dying (keep in mind this is just an example table, I have about 30,000 rows so nothing too crazy) 这是我写的代码,但是内核一直在死(请记住这只是一个示例表,我有大约30,000行,所以没什么太疯狂的)

df['C'] = df.loc[df['B'] > 0, 'A'] / df['B'])

any idea on what's going on? 对于发生了什么的任何想法? Is something running infinitely that's causing it to crash? 是无限运行会导致它崩溃吗? Thanks for the help. 谢谢您的帮助。

You get that using np.where 你可以使用np.where获得它

df['C'] = np.round(np.where(df['B'] > 0, df['A']/df['B'], 0), 1)

Or if you want to use loc 或者如果你想使用loc

df.loc[df['B'] > 0, 'C'] = df['A']/df['B']

and then fillna(0) 然后fillna(0)

Option 1 选项1
You use pd.Series.mask to hide zeros, and then just empty cells with fillna . 使用pd.Series.mask隐藏零,然后使用pd.Series.mask清空单元fillna

v = (df.A / df.B.mask(df.B == 0)).fillna(0)
v

0     0.000000
1     0.000000
2     2.500000
3     4.000000
4     1.000000
5     0.000000
6     1.000000
7     1.000000
8     0.333333
9     0.000000
10    1.666667
11    0.000000
dtype: float64

df['C'] = v

Alternatively, replace those zeros with np.inf , because x / inf = 0 . 或者,用np.inf替换那些零,因为x / inf = 0

df['C'] = (df.A / df.B.mask(df.B == 0, np.inf))

Option 2 选项2
Direct replacement with df.replace df.replace直接替换

df.A / df.B.replace(0, np.inf)

0     0.000000
1     0.000000
2     2.500000
3     4.000000
4     1.000000
5     0.000000
6     1.000000
7     1.000000
8     0.333333
9     0.000000
10    1.666667
11    0.000000
dtype: float64

Keep in mind, you can do an astype conversion, if you want mixed integers and floats as your result: 请记住,如果你想要混合整数和浮点数,你可以做一个astype转换:

df.A.div(df.B.replace(0, np.inf)).astype(object)

0            0
1            0
2          2.5
3            4
4            1
5            0
6            1
7            1
8     0.333333
9            0
10     1.66667
11           0
dtype: object

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