Question

我有一个如下数据框。我正在尝试检查nan中是否有Liq_Factor，如果可以，请放入1，否则除以use/TW。结果进入“测试”列。

+---+------------+------------+--------+--------+--------+
| 1 |            | Liq_Factor | Zscire | Use    | Tw     |
| 2 | 01/10/2020 | 36.5       | 44     | 43.875 | 11.625 |
| 3 | 02/10/2020 | Nan        | 43.625 | 13.625 | 33.25  |
| 4 | 03/10/2020 | 6.125      | 47.875 | 22.5   | 4.625  |
| 5 | 04/10/2020 | Nan        | 34.25  | 37.125 | 36     |
| 6 | 05/10/2020 | 43.875     | 17.375 | 5.5    | 36.25  |
| 7 | 06/10/2020 | 40         | 14.125 | 21.125 | 14.875 |
| 8 | 07/10/2020 | 42.25      | 44.75  | 21.25  | 31.75  |
+---+------------+------------+--------+--------+--------+

我想知道我是否可以使用.apply

DF1['Testing']=(DF1['Liq_Factor'].apply(lambda x: x=1 if pd.isna(DF1['Zscore']) else DF1['Use']/DF1['Tw'])

可以帮忙吗？

谢谢，高

Answer 1

您可以使用apply或另一种替代方法，其中numpy中的函数是

df['Liq_Factor'] = np.where(df['Liq_Factor'] == np.Nan, 1, df['Use']/df['TW'])

您可以在下面的评论中进行以下操作：

# create another column with the calculation
df['calc'] = (1/3)* df['ATV']/df['TW']*100000000
# create two rules (you can use one rule and then the opposite)
mask_0 = (df['calc'] < 1)
mask_1 = (df['calc'] > 1)
# change result value by condition
df.loc[mask_0, 'Liq Factor'] = df['calc']
df.loc[mask_1, 'Liq Factor'] = 1

Answer 2

使用以下代码-

df['Testing']=df.apply(lambda x: 1 if x['Liq_Factor']=='Nan'  else x['Use']/x['Tw'], axis=1)

基于评论部分的更改

df['Testing']=df.apply(lambda x: 1 if x['Liq_Factor']=='Nan'  else min(x['Use']/x['Tw'],1), axis=1)

如果在不同的列中，熊猫有条件地申请

2 个答案: