您需要
eq使用
shift：
df['match'] = df.col1.eq(df.col1.shift())print (df) col1 match0 1 False1 3 False2 3 True3 1 False4 2 False5 3 False6 2 False7 2 True
或改为
eq使用
==，但是在大型Dataframe中，它会稍微慢一些：
df['match'] = df.col1 == df.col1.shift()print (df) col1 match0 1 False1 3 False2 3 True3 1 False4 2 False5 3 False6 2 False7 2 True
时间 ：
import pandas as pddata={'col1':[1,3,3,1,2,3,2,2]}df=pd.Dataframe(data,columns=['col1'])print (df)#[80000 rows x 1 columns]df = pd.concat([df]*10000).reset_index(drop=True)df['match'] = df.col1 == df.col1.shift()df['match1'] = df.col1.eq(df.col1.shift())print (df)In [208]: %timeit df.col1.eq(df.col1.shift())The slowest run took 4.83 times longer than the fastest. This could mean that an intermediate result is being cached.1000 loops, best of 3: 933 µs per loopIn [209]: %timeit df.col1 == df.col1.shift()1000 loops, best of 3: 1 ms per loop
