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        Pandas:用于在 DataFrame 中设置值的三元条件运算符

        Pandas: Ternary conditional operator for setting a value in a DataFrame(Pandas:用于在 DataFrame 中设置值的三元条件运算符)
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                  本文介绍了Pandas:用于在 DataFrame 中设置值的三元条件运算符的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

                  问题描述

                  我有一个数据框 pd.我想更改列 irr 的值,具体取决于它是高于还是低于阈值.

                  I have a dataframe pd. I would like to change a value of column irr depending on whether it is above or below a thresh hold.

                  我怎样才能在一行中做到这一点?现在我有

                  How can I do this in a single line? Now I have

                  pd['irr'] = pd['irr'][pd['cs']*0.63 > pd['irr']] = 1.0
                  pd['irr'] = pd['irr'][pd['cs']*0.63 <=  pd['irr']] = 0.0
                  

                  问题当然是我改了irr,在下一行再次检查.

                  The problem of course is that I change irr and check it again in the next line.

                  是否有类似 pandas 的三元条件运算符?

                  Is there something like a ternary conditional operator for pandas?

                  推荐答案

                  在 pandas 中没有,在 numpy 中是.

                  In pandas no, in numpy yes.

                  您可以使用 numpy.where 或将条件创建的 boolean Series 转换为 float - True1.0>Falses 是 0.0:

                  You can use numpy.where or convert boolean Series created by condition to float - Trues are 1.0 and Falses are 0.0:

                  pd['irr'] = np.where(pd['cs']*0.63 > pd['irr'], 1.0, 0.0)
                  

                  或者:

                  pd['irr'] = (pd['cs']*0.63 > pd['irr']).astype(float)
                  

                  示例:

                  pd = pd.DataFrame({'cs':[1,2,5],
                                     'irr':[0,100,0.04]})
                  
                  print (pd)
                     cs     irr
                  0   1    0.00
                  1   2  100.00
                  2   5    0.04
                  
                  pd['irr'] = (pd['cs']*0.63 > pd['irr']).astype(float)
                  print (pd)
                     cs  irr
                  0   1  1.0
                  1   2  0.0
                  2   5  1.0
                  

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