python - 当为另一列指定一个条件时如何填充前向值？

Question

我正在分析如下一张表。 offer_id 中有几个 None 值当事件列显示 transaction 时的列。仅当前一个事件是 offer viewed 时，我才会通过值forward 填充 None ，否则，将 None 值填充为 0 或将其保留为 none。

数据框是:

df = pd.DataFrame({'event': ['offer_received', 'offer_viewed','transaction', 'transaction', 'offer_received', 'transaction'], 'user':['A','A','A','A','A','A'], 'value':[0, 0, 1.09, 2.55, 0, 3.02], 'offer_id': ['0b1e1539f2cc45b7b9fa7c272da2e1d7', '0b1e1539f2cc45b7b9fa7c272da2e1d7', 'None', 'None', '3f207df678b143eea3cee63160fa8bed', 'None'], 'days':[0, 0.25, 9.75, 11, 0,9.75]})

event           user   value    offer_id                            days
offer received  A      0.00     0b1e1539f2cc45b7b9fa7c272da2e1d7    0.00
offer viewed    A      0.00     0b1e1539f2cc45b7b9fa7c272da2e1d7    0.25
transaction     A      1.09     None                                9.75
transaction     A      2.55     None                                11
offer received  A      0.00     3f207df678b143eea3cee63160fa8bed    0.00
transaction     A      3,02     None                                9.75

我尝试使用df.offer_id.fillna(method = 'ffill')但我只是不知道当上一个事件是 offer_viewed 时如何将条件放在事件列上然后填写offer_id的transaction通过使用(method = 'ffill') .

我的预期结果是这样的:

event           user   value    offer_id                            days
offer received  A      0.00     0b1e1539f2cc45b7b9fa7c272da2e1d7    0.00
offer viewed    A      0.00     0b1e1539f2cc45b7b9fa7c272da2e1d7    0.2
transaction     A      1.09     0b1e1539f2cc45b7b9fa7c272da2e1d7    9.75                   
transaction     A      2.55     0b1e1539f2cc45b7b9fa7c272da2e1d7    11                         
offer received  A      0.00     3f207df678b143eea3cee63160fa8bed    0.00
transaction     A      3,02     None                                9.75

Answer 1

我认为你可以使用 shift()、ffill() 和 where() 来实现:

df = pd.DataFrame({'e': ['r', 'v', 't', 'r', 't'], 'oid': [1, 1, np.nan, 2, np.nan]})
df
#    e  oid
# 0  r  1.0
# 1  v  1.0
# 2  t  NaN
# 3  r  2.0
# 4  t  NaN
df.oid = df.oid.ffill().where(df.e.shift() == 'v', df.oid)
df
#    e  oid
# 0  r  1.0
# 1  v  1.0
# 2  t  1.0
# 3  r  2.0
# 4  t  NaN

您甚至可以跳过 ffill() 并使用 shift() 两次:

df = pd.DataFrame({'e': ['r', 'v', 't', 'r', 't'], 'oid': [1, 1, np.nan, 2, np.nan]})

df.oid = df.oid.shift().where(df.e.shift() == 'v', df.oid)
df
#    e  oid
# 0  r  1.0
# 1  v  1.0
# 2  t  1.0
# 3  r  2.0
# 4  t  NaN