我有两个DataFrame Multi-indexed。 一个是我的引用(大约37000行),另一个是包含较少的值(例如ex 10)
我想用第二个值替换大的行
例如:df1
:(很抱歉,显示效果不佳)
lvl1 lvl2 lvl3 Value Value2
A 1 I 0,862877333 0,181795348
1 II 0,787022218 0,292046262
1 III 0,40516176 0,445079108
2 I 0,882167166 0,683954412
2 IV 0,743618024 0,103097267
3 I 0,901062673 0,729188996
3 II 0,529989452 0,715379923
3 IV 0,740272198 0,792457421
B 1 I 0,548587694 0,637462653
1 II 0,201284924 0,084391963
2 I 0,999118031 0,558207224
2 II 0,63353019 0,251377184
2 V 0,694294638 0,685050861
3 V 0,436723389 0,310871641
3 VI 0,630832871 0,869957421
3 VII 0,157874482 0,639308814
df 2
lvl1 lvl2 lvl3 Value Value2
A 1 I 0,8654 1
B 2 II 0,264 2
产生的df3
:
lvl1 lvl2 lvl3 Value Value2
A 1 I **0,8654** 0,181795348
1 II 0,787022218 0,292046262
1 III 0,40516176 0,445079108
2 I 0,882167166 0,683954412
2 IV 0,743618024 0,103097267
3 I 0,901062673 0,729188996
3 II 0,529989452 0,715379923
3 IV 0,740272198 0,792457421
1 I 0,548587694 0,637462653
B 1 II 0,201284924 0,08439196
2 I 0,999118031 0,558207224
2 II **0,264** 0,251377184
2 V 0,694294638 0,685050861
3 V 0,436723389 0,310871641
3 VI 0,630832871 0,869957421
3 VII 0,157874482 0,639308814
答案 0 :(得分:0)
您也许可以使用pd.merge
import numpy as np
import pandas as pd
temp = pd.DataFrame({"lvl1": ["A","A","B","B"], "lvl2": [1,2,1,2], "lvl3": ["I","II","I","II"], "Value": [0.8628773,0.7870, 0.63353, 0.6998]})
replace = pd.DataFrame({"lvl1": ["A","B"], "lvl2": [1,2], "lvl3": ["I","II"], "Value": [0.8654, 0.264], "Value2": [1,2]})
df = pd.merge(temp, replace, how="left", on=["lvl1","lvl2","lvl3"])
df["Value_x"] = np.where(df["Value_y"].notnull(), df["Value_y"], df["Value_x"])
# df.drop(["Value_y", "Value2"], axis=1, inplace=True)
答案 1 :(得分:0)
您可以尝试像这样替换索引匹配中的值:
for ind in df2.index:
df1.loc[ind, 'Value'] = df2.loc[ind, 'Value']
如果您要替换行:
for ind in df2.index:
df1.loc[ind,] = df2.loc[ind,]