您如何找到数据框中每一行的值何时发生变化？

Question

我正在尝试将帐户标记为新帐户、当前帐户、丢失帐户或返回帐户，但我遇到了逻辑问题。行索引是帐户，列是年份，值是 1 和 0，表示帐户是否处于活动状态。这是我到目前为止想出的。我不确定这是否会奏效，或者我是否已经接近，我不确定逻辑将如何寻找回头客。 df2 是原始数据框，df3 = df2.shift(periods=1,axis=1)

def differences():
    if df2 != df3 & df2 == 1:
        return "New"
    elif df2 != df3 & df2 ==0:
        return "Lost"
    elif df2 == df3 & df2 ==0:
        return ""
    else:
        return "Continuing"
differences() 

`

当我运行此代码时，我收到以下错误：

couldn't find matching opcode for 'and_bdl'

Answer 1

以下代码逻辑可能适用于您的情况。

编辑：根据您的评论，我修改了代码，以便检查除最后​​一列之外的所有列。

import pandas as pd

str="""account  2019  2018  2017  2016  2015
alex 1 0 0 0 0
joe  0 0 1 0 0
boss  1 1 1 1 1
smith 1 1 0 1 0"""
df = pd.read_csv(pd.io.common.StringIO(str), sep='\s+', index_col='account')
df
#Out[46]: 
#         2019  2018  2017  2016  2015
#account                              
#alex        1     0     0     0     0
#joe         0     0     1     0     0
#boss        1     1     1     1     1
#smith       1     1     0     1     0

# find account status per-year
def account_status(x):
    status = []
    n = x.size
    for i in range(n-1):
        if x.iloc[i] == 1: 
            # if all rest are '0'
            if x.iloc[i+1:].eq(0).all():
                status.extend(['new'] + [None]*(n-i-2))
                break            
            # if the previous year is '0'
            elif x.iloc[i+1] == 0:
                status.append('returning')
            else:
                status.append('continuing')
        else:
            # at least one '1' in previous years
            if x.iloc[i+1:].eq(1).any():
                status.append('lost')
            else:
                status.extend([None] * (n-i-1))
                break
    return status    

s = df.apply(account_status, axis=1).apply(pd.Series)
s.columns = df.columns[:-1]
s                                                                                                                   
#Out[57]: 
#               2019        2018        2017        2016
#account                                                
#alex            new        None        None        None
#joe            lost        lost         new        None
#boss     continuing  continuing  continuing  continuing
#smith    continuing   returning        lost         new