Python Pandas Dataframe用列表中的值替换NaN

Question

我正在尝试用NaN代替我的专栏

group_choices = ['Group1', 'Group2', 'Group3']

Groups limit
1 NaN NaN
2 Group1 2
3 Group2 2
4 Group3 2
5 NaN NaN
6 NaN NaN
7 NaN NaN

如何根据group_choises随机替换NaN？

由于限制列中的限制值，我还试图限制可以随机选择group_choise的频率。

我正在尝试获得以下结果：

Groups limit
1 Group3 NaN
2 Group1 2
3 Group2 2
4 Group3 2
5 Group1 NaN
6 Group2 NaN
7 Out of groups

Answer 1

fillna和字典

dct = dict(zip(df.Groups.loc[pd.isna].index, group_choices))

df.fillna({'Groups': dct}).fillna({'Groups': 'Out of groups'})

          Groups  limit
1         Group1    NaN
2         Group1    2.0
3         Group2    2.0
4         Group3    2.0
5         Group2    NaN
6         Group3    NaN
7  Out of groups    NaN

---

旧答案

有用，但我更喜欢新的。它说明了我思考过程的演变。

发电机

def get_some(i, n):
  for x in [*i] * n:
    yield x

def fill(s, i, n):
  gs = get_some(i, n)
  for x in s:
    if pd.isnull(x):
      try:
        yield next(gs)
      except StopIteration:
        yield "Out of groups"
    else:
      yield x

df.assign(Groups=[*fill(df.Groups, group_choices, 1)])

          Groups  limit
1         Group1    NaN
2         Group1    2.0
3         Group2    2.0
4         Group3    2.0
5         Group2    NaN
6         Group3    NaN
7  Out of groups    NaN

---

替代

def get_some(i, n):
  for x in [*i] * n:
    yield x

df.assign(Groups=df.Groups.fillna(
    df.Groups.loc[pd.isna].pipe(
        lambda s: pd.Series(dict(zip(s.index, get_some(group_choices, 1))))
    )
).fillna('Out of groups'))