Python 循环浏览每个客户记录，以获取他们来自的第一个最后一个渠道

1. 将Date转换为DateTime，以便可以按Date排序(如果日期采用此格式，则使用format="%m/%d/%y").
2. 定义condition_i和condition_ii来解释你的新逻辑:你想保留for each 102:(i)如果所有行的Channel都是Direct，则保留all rows；或者(ii)如果所有行的Channel都是Direct，则保留only rows where 103 is 109 104.
3. 分组为Customer ID，然后分别由first和last相加.
4. 合并由组比产生的帧.

df["Date"] = pd.to_datetime(df["Date"], format="%d/%m/%y")

df = df.sort_values(by="Date")

condition_i = df.groupby("Customer ID")["Channel"].transform(
    lambda x: all(x == "Direct")
)

condition_ii = (
    df.groupby("Customer ID")["Channel"]
    .transform(lambda x: any(x != "Direct"))
    .loc[df["Channel"] != "Direct"]
)

first = (
    df[(condition_i) | (condition_ii)]
    .groupby("Customer ID", as_index=False)
    .first()
    .rename(columns={"Channel": "Fist Channel"})
)

last = (
    df[(condition_i) | (condition_ii)]
    .groupby("Customer ID", as_index=False)
    .last()
    .rename(columns={"Channel": "Last Channel"})
)

out = pd.merge(
    first[["Customer ID", "Fist Channel"]], last[["Customer ID", "Last Channel"]]
)

   Customer ID Fist Channel Last Channel
0            1        Email       Search
1            2         Paid        Email
2            3       Direct       Direct