1
我想用字典中的值替換數據幀的值。 In English簡體中文:如果Column C
中的某個值與字典密鑰匹配,則用與該特定密鑰對應的字典中的值替換Column D
。基於詞典設置熊貓值
import pandas as pd
import numpy as np
dfp = pd.DataFrame({'A' : [np.NaN,np.NaN,3,4,5,5,3,1,5,np.NaN],
'B' : [1,0,3,5,0,0,np.NaN,9,0,0],
'C' : ['AA1233445','A9875', 'rmacy','Idaho Rx','Ab123455','TV192837','RX','Ohio Drugs','RX12345','USA Pharma'],
'D' : [123456,123456,1234567,12345678,12345,12345,12345678,123456789,1234567,np.NaN],
'E' : ['Assign','Unassign','Assign','Ugly','Appreciate','Undo','Assign','Unicycle','Assign','Unicorn',]})
print(dfp)
z = {'rmacy': 999}
dfp.loc[dfp['C'].isin(z.keys()), 'D' ] = z.values() # <--- code to change
Output:
A B C D E
0 NaN 1.0 AA1233445 123456 Assign
1 NaN 0.0 A9875 123456 Unassign
2 3.0 3.0 rmacy (999) Assign #<--- Worked with paranthesis
3 4.0 5.0 Idaho Rx 1.23457e+07 Ugly
4 5.0 0.0 Ab123455 12345 Appreciate
5 5.0 0.0 TV192837 12345 Undo
6 3.0 NaN RX 1.23457e+07 Assign
7 1.0 9.0 Ohio Drugs 1.23457e+08 Unicycle
8 5.0 0.0 RX12345 1.23457e+06 Assign
9 NaN 0.0 USA Pharma NaN Unicorn
上面的代碼工作(除非是放入Paranthesis的價值。但如果字典是大於一個鍵,就會把兩個值中Column D
因爲有在列兩場比賽。
A B C D E
0 NaN 1.0 AA1233445 123456 Assign
1 NaN 0.0 A9875 123456 Unassign
2 3.0 3.0 rmacy (999, 333) Assign
3 4.0 5.0 Idaho Rx 1.23457e+07 Ugly
4 5.0 0.0 Ab123455 12345 Appreciate
5 5.0 0.0 TV192837 12345 Undo
6 3.0 NaN RX (999, 333) Assign
7 1.0 9.0 Ohio Drugs 1.23457e+08 Unicycle
8 5.0 0.0 RX12345 1.23457e+06 Assign
9 NaN 0.0 USA Pharma NaN Unicorn
一個將如何解決這個問題?
你總是給出正確答案!哈哈。你能通過這個方法走過我嗎?如何使用'fillna()'給出正確的結果正在引起我的注意。 – MattR
@MattR'map'返回字典中的值(如果有),否則返回'NaN'。然後我用'fillna(df.D)'填充這些'NaN'。然後我用這些新值覆蓋當前的'df.D'。 – piRSquared
簡直太棒了......我希望我有一天能像你一樣流利。乾杯! – MattR