对于此表:
我想生成“ desired_output”列。一种实现此目的的方法可能是:
我尝试过的代码:
df['desired_output']=df.col_1.apply(lambda x: True if x.shift()==True else False)
谢谢
答案 0 :(得分:5)
您可以按|
链接按位OR
的原始d = {"col1":[False,True,True,True,False,True,False,False,True,False,False,False]}
df = pd.DataFrame(d)
df['new'] = df.col1 | df.col1.shift(-1)
print (df)
col1 new
0 False True
1 True True
2 True True
3 True True
4 False True
5 True True
6 False False
7 False True
8 True True
9 False False
10 False False
11 False False
,其值偏移Series.shift
:
7-Zip 18.05 (x64) : Copyright (c) 1999-2018 Igor Pavlov : 2018-04-30
Scanning the drive for archives:
1 file, 40021368 bytes (39 MiB)
Testing archive: C:\Users\Lozzy\Documents\ARDF\broken\Cart_Weel_#10-AT_2020-08-06_13268.txt.gz
--
Path = C:\Users\Lozzy\Documents\ARDF\broken\Cart_Weel_#10-AT_2020-08-06_13268.txt.gz
Type = gzip
Headers Size = 10
Sub items Errors: 1
Archives with Errors: 1
Sub items Errors: 1
7-Zip 18.05 (x64) : Copyright (c) 1999-2018 Igor Pavlov : 2018-04-30
Scanning the drive for archives:
1 file, 40021368 bytes (39 MiB)
Testing archive: C:\Users\Lozzy\Documents\ARDF\broken\Cart_Weel_#210-AT_2020-08-06_13268.txt - Copy.gz
--
Path = C:\Users\Lozzy\Documents\ARDF\broken\Cart_Weel_#210-AT_2020-08-06_13268.txt - Copy.gz
Type = gzip
Headers Size = 10
Sub items Errors: 1
Archives with Errors: 1
Sub items Errors: 1
7-Zip 18.05 (x64) : Copyright (c) 1999-2018 Igor Pavlov : 2018-04-30
Scanning the drive for archives:
1 file, 56581 bytes (56 KiB)
Testing archive: C:\Users\Lozzy\Documents\ARDF\broken\Cart_Weel_#DT-F@_2020-08-06_13268.txt.gz
--
Path = C:\Users\Lozzy\Documents\ARDF\broken\Cart_Weel_#DT-F@_2020-08-06_13268.txt.gz
Type = gzip
Headers Size = 10
Everything is Ok
Size: 504716
Compressed: 56581
答案 1 :(得分:2)
尝试
df['desired_output'] = df['col_1']
df.loc[1:, 'desired_output'] = df.col_1[1:].values | df.col_1[:-1].values
print(df)
答案 2 :(得分:1)
如果这些另存为字符串。 all_caps(是/否) 输入:
col_1
0 True
1 True
2 False
3 True
4 True
5 False
6 Flase
7 True
8 False
代码:
df['desired']=df['col_1']
for i, e in enumerate(df['col_1']):
if e=='True':
df.at[i-1,'desired']=df.at[i,'col_1']
df = df[:(len(df)-1)]
df
输出:
col_1 desired
0 True True
1 True True
2 False True
3 True True
4 True True
5 False False
6 Flase True
7 True True
8 False False