Question

我正在尝试根据收到的报告制作非规范化数据框。我需要将记录分配给一个组，该组来自一行，每组之间有随机文本和nan。满足条件时如何重写这些行值？我编写的循环似乎仅在满足条件时才覆盖下一个值，直到满足下一个条件时才执行此操作。请在下面查看我的数据和代码示例。本质上，我需要将行设置为主行，辅助行或我决定的任何其他组，但是行必须一直运行到命中下一个指定组为止。

当前数据：

    Primary
    Week#
    1
    nan
    nan
    nan
    2
    nan
    nan
    nan
    Secondary
    Week#
    1
    nan
    nan
    nan
    2
    nan
    nan
    nan

代码：

for index, obj in enumerate(df['col0']):
    l = len(df['col0'])
    if obj == 'Primary':
        if index > 0:
            previous = df['col0'][index - 1]
        if index < (l - 1):
            next_ = df['col0'][index + 1]
            next_ = obj
            print (next_, obj)

    if obj == 'Secondary':
        if index > 0:
            previous = df['col0'][index - 1]
        if index < (l - 1):
            next_ = df['col0'][index + 1]
            next_ = obj
            print (next_, obj)

预期输出：

Primary
Primary
Primary
Primary
Primary
Primary
Primary
Primary
Primary
Primary
Secondary
Secondary
Secondary
Secondary
Secondary
Secondary
Secondary
Secondary
Secondary
Secondary

Answer 1

执行此操作的更紧急的方法是仅保留您关心的值，然后将其填充到不需要的地方。例如：

df["col0_cleaned"] = df["col0"].where(df["col0"].isin(["Primary", "Secondary"])).ffill()

如果我们逐步采取这一措施，将会更加清楚发生了什么事情：

df["isin"] = df["col0"].isin(["Primary", "Secondary"])
df["where"] = df["col0"].where(df["col0"].isin(["Primary", "Secondary"]))
df["ffill"] = df["col0"].where(df["col0"].isin(["Primary", "Secondary"])).ffill()

这给了我

In [350]: df
Out[350]: 
         col0   isin      where      ffill
0     Primary   True    Primary    Primary
1       Week#  False        NaN    Primary
2           1  False        NaN    Primary
3         nan  False        NaN    Primary
4         nan  False        NaN    Primary
5         nan  False        NaN    Primary
6           2  False        NaN    Primary
7         nan  False        NaN    Primary
8         nan  False        NaN    Primary
9         nan  False        NaN    Primary
10  Secondary   True  Secondary  Secondary
11      Week#  False        NaN  Secondary
12          1  False        NaN  Secondary
13        nan  False        NaN  Secondary
14        nan  False        NaN  Secondary
15        nan  False        NaN  Secondary
16          2  False        NaN  Secondary
17        nan  False        NaN  Secondary
18        nan  False        NaN  Secondary
19        nan  False        NaN  Secondary

满足条件时如何用以前的值替换值？

1 个答案: