我已经设置了以下数据框
A
[ClaimantId], [ClaimId], [LenderId], [IsWorked]
1 1 1 1
1 2 4 0
1 3 3 1
2 6 1 1
B
[ClaimantId], [Forename], [Surname]
1 Bruce Wayne
2 Peter Parker
我想要的输出为
[ClaimantId], [Forename], [Surname], [C1], [C2], [C3], [L1], [L2], [L3], [W1], [W2], [W3]
1 Bruce Wayne 1 2 3 1 4 3 1 0 1
2 Peter Parker 6 Nan Nan 1 Nan Nan 1 Nan Nan
我不确定该如何应用,C / L / W列数的上限为20,并且永远不会超过。
非常感谢您的帮助。
谢谢
答案 0 :(得分:1)
使用:
d = {'ClaimId':'C', 'LenderId':'L','IsWorked':'W'}
df = (A.rename(columns=d)
.set_index(['ClaimantId',A.groupby('ClaimantId').cumcount()])
.unstack())
df.columns = [f'{i}{j+1}' for i, j in df.columns]
print (df)
C1 C2 C3 L1 L2 L3 W1 W2 W3
ClaimantId
1 1.0 2.0 3.0 1.0 4.0 3.0 1.0 0.0 1.0
2 6.0 NaN NaN 1.0 NaN NaN 1.0 NaN NaN
df1 = B.join(df, on='ClaimantId')
print (df1)
ClaimantId Forename Surname C1 C2 C3 L1 L2 L3 W1 W2 \
0 1 Bruce Wayne 1.0 2.0 3.0 1.0 4.0 3.0 1.0 0.0
1 2 Peter Parker 6.0 NaN NaN 1.0 NaN NaN 1.0 NaN
W3
0 1.0
1 NaN
说明:
rename
列set_index
创建的计数器Series
cumcount
unstack
重塑list comprehension
与f-string
s一起平铺MultiIndex列join
秒DataFrame
编辑:
如果需要相同的长度,则所有列将由MultiIndex
创建的新range
使用reindex
:
d = {'ClaimId':'C', 'LenderId':'L','IsWorked':'W'}
df = (A.rename(columns=d)
.set_index(['ClaimantId',A.groupby('ClaimantId').cumcount()])
.unstack())
mux = pd.MultiIndex.from_product([df.columns.get_level_values(0).unique(), range(5)])
df = df.reindex(columns=mux, fill_value=0)
df.columns = [f'{i}{j+1}' for i, j in df.columns]
print (df)
C1 C2 C3 C4 C5 L1 L2 L3 L4 L5 W1 W2 W3 W4 \
ClaimantId
1 1.0 2.0 3.0 0 0 1.0 4.0 3.0 0 0 1.0 0.0 1.0 0
2 6.0 NaN NaN 0 0 1.0 NaN NaN 0 0 1.0 NaN NaN 0
W5
ClaimantId
1 0
2 0