我有一個數據框如下。nan我正在嘗試檢查中是否有 a Liq_Factor，如果是，則將 1 放入否則除法use/TW。結果列于測試中。+---+------------+------------+--------+--------+--------+| 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.apply使用DF1['Testing']=(DF1['Liq_Factor'].apply(lambda x: x=1 if pd.isna(DF1['Zscore']) else DF1['Use']/DF1['Tw'])你能幫忙嗎？
2 回答
蝴蝶不菲
TA貢獻1810條經驗 獲得超4個贊
您可以使用 apply 或另一種替代方法是 numpy 中的 where 函數：
df['Liq_Factor'] = np.where(df['Liq_Factor'] == np.Nan, 1, df['Use']/df['TW'])
按照下面的評論，您可以執行以下操作：
# 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
楊魅力
TA貢獻1811條經驗 獲得超6個贊
使用下面的代碼-
df['Testing']=df.apply(lambda x: 1 if x['Liq_Factor']=='Nan' else x['Use']/x['Tw'], axis=1)
根據評論部分的更改
df['Testing']=df.apply(lambda x: 1 if x['Liq_Factor']=='Nan' else min(x['Use']/x['Tw'],1), axis=1)
添加回答
舉報
0/150
提交
取消