列中重复值之间的条件答案

【问题标题】：Condition between duplicated values in a column列中重复值之间的条件
【发布时间】：2019-03-27 11:06:47
【问题描述】：

当每个客户有多个计划时，他们就会被重复。我想给客户设置状态：

如果他们每个产品都填写了“canceled_at”，则客户状态被取消，但当不是所有产品都填写了“cancelled_at”，而是至少有一个产品时，状态为“降级”，因为他丢失了一个产品。

customer|canceled_at|status
x       |3/27/2018  |
x       |           |
y       |2/2/2018   |
y       |2/2/2018   |
z       |1/1/2018   |
a       |           |

我已经有取消状态，现在只需要降级

df['status']=(df.groupby('customer')['canceled_at'].
  transform(lambda x: x.notna().all()).map({True:'canceled'})).fillna(df.status)

customer|canceled_at|status
x       |3/27/2018  |downgrade
x       |           |downgrade
y       |2/2/2018   |canceled
y       |2/2/2018   |canceled
z       |1/1/2018   |canceled
a       |           |

【问题讨论】：

标签： python pandas pandas-groupby pandas-loc

【解决方案1】：

这里可以比较列是否没有缺失值，并按Series customer 与GroupBy.transform 和GroupBy.all 分组， GroupBy.any 用于测试所有值 Trues（所有未丢失）或至少一个未丢失值（任何未丢失）并将其传递给 numpy.select：

g = df['canceled_at'].notna().groupby(df['customer'])
m1 = g.transform('all')
m2 = g.transform('any')

df['status'] = np.select([m1, m2],['canceled','downgrade'], np.nan)
print (df)
  customer canceled_at     status
0        x   3/27/2018  downgrade
1        x         NaN  downgrade
2        y    2/2/2018   canceled
3        y    2/2/2018   canceled
4        z    1/1/2018   canceled
5        a         NaN        nan

或者：

df['status'] = np.select([m1, m2],['canceled','downgrade'], '')
print (df)
  customer canceled_at     status
0        x   3/27/2018  downgrade
1        x         NaN  downgrade
2        y    2/2/2018   canceled
3        y    2/2/2018   canceled
4        z    1/1/2018   canceled
5        a         NaN

如果只有NaNs 组需要转换为downgrade：

mask = df['canceled_at'].notna().groupby(df['customer']).transform('all')
df['status'] = np.where(mask,'canceled','downgrade')
print (df)
  customer canceled_at     status
0        x   3/27/2018  downgrade
1        x         NaN  downgrade
2        y    2/2/2018   canceled
3        y    2/2/2018   canceled
4        z    1/1/2018   canceled
5        a         NaN  downgrade

【讨论】：

【解决方案2】：

这是一种方法：

import pandas as pd

def select_status(canceled):
    c = canceled.count()
    if c == 0:
        status = ''
    elif c == len(canceled):
        status = 'canceled'
    else:
        status = 'downgrade'
    return pd.Series(status, index=canceled.index)

df = pd.DataFrame({'customer': ['x', 'x', 'y', 'y', 'z', 'a'],
                   'canceled_at': ['3/27/2018', None, '2/2/2018', '2/2/2018', '1/1/2018', None]})
df['status'] = df.groupby('customer')['canceled_at'].apply(select_status)
print(df)

输出：

  customer canceled_at     status
0        x   3/27/2018  downgrade
1        x        None  downgrade
2        y    2/2/2018   canceled
3        y    2/2/2018   canceled
4        z    1/1/2018   canceled
5        a        None

【讨论】：