我想找出 pandas DataFrame 中两列 int 类型的区别。我正在使用 python 2.7。列如下 -
>>> df
INVOICED_QUANTITY QUANTITY_SHIPPED
0 15 NaN
1 20 NaN
2 7 NaN
3 7 NaN
4 7 NaN
现在,我想从 INVOICED_QUANTITY 中减去 QUANTITY_SHIPPED,然后执行以下操作-
>>> df['Diff'] = df['QUANTITY_INVOICED'] - df['SHIPPED_QUANTITY']
>>> df
QUANTITY_INVOICED SHIPPED_QUANTITY Diff
0 15 NaN NaN
1 20 NaN NaN
2 7 NaN NaN
3 7 NaN NaN
4 7 NaN NaN
如何处理 NaN?我希望得到以下结果,因为我希望将 NaN 视为 0(零)-
>>> df
QUANTITY_INVOICED SHIPPED_QUANTITY Diff
0 15 NaN 15
1 20 NaN 20
2 7 NaN 7
3 7 NaN 7
4 7 NaN 7
我不想做 df.fillna(0)
。总而言之,我会尝试类似以下的方法并且它有效但没有区别 -
>>> df['Sum'] = df[['QUANTITY_INVOICED', 'SHIPPED_QUANTITY']].sum(axis=1)
>>> df
INVOICED_QUANTITY QUANTITY_SHIPPED Diff Sum
0 15 NaN NaN 15
1 20 NaN NaN 20
2 7 NaN NaN 7
3 7 NaN NaN 7
4 7 NaN NaN 7
最佳答案
您可以使用 sub
方法执行减法 - 此方法允许将 NaN
值视为指定值:
df['Diff'] = df['INVOICED_QUANTITY'].sub(df['QUANTITY_SHIPPED'], fill_value=0)
产生:
INVOICED_QUANTITY QUANTITY_SHIPPED Diff
0 15 NaN 15
1 20 NaN 20
2 7 NaN 7
3 7 NaN 7
4 7 NaN 7
另一个巧妙的方法是 @JianxunLi suggests : 填写列中的缺失值(创建列的副本)并照常减去。
这两种方法几乎相同,尽管 sub
效率更高一些,因为它不需要提前生成列的副本;它只是“即时”填充缺失值:
In [46]: %timeit df['INVOICED_QUANTITY'] - df['QUANTITY_SHIPPED'].fillna(0)
10000 loops, best of 3: 144 µs per loop
In [47]: %timeit df['INVOICED_QUANTITY'].sub(df['QUANTITY_SHIPPED'], fill_value=0)
10000 loops, best of 3: 81.7 µs per loop
关于python - 使用 Pandas 查找具有 Null 的 2 列之间的差异,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31053848/