用列python中的新整数替换所有非整数

Question

所以我有一个名为所有者的列，看起来像这样

'Owned by Terrick Inc.',
 nan,
 'Owned by MHPI Inc.',
 'Owned by Zeman Homes & Neighborhoods',
 'Owned by Santefort Neighborhood Properties',
 'Owned by MHPI Inc.',
 'Owned by MHPI Inc.',
 nan,
 2,
 nan,
 5,
 'Owned by RHP Properties',
 'Owned by Zeman Homes & Neighborhoods',
 'Owned by Zeman Homes & Neighborhoods',
 'Owned by Gateway Manufactured Home Community',
 3,
 nan,
 'Owned by Zeman Homes & Neighborhoods',
 'Owned by MHPI Inc.',
 nan,
 'Owned by Indian Trails LLC',
 'Owned by Zeman Homes & Neighborhoods',
 1,
 'Owned by Lake View Manufactured Home Community',
 5,
 2,
 'Owned by Zeman Homes & Neighborhoods'

我正在尝试将所有不是数字的值更改为数字 10。但由于某种原因它无法正常工作。我尝试了 2 种方法，这是第一种，但这只是将整个列替换为 9，因此它不符合我的条件。

df['Owner'][(df['Owner'] != 0) | (df['Owner'] != '1') | (df['Owner'] != '2') | (df['Owner'] != 3) | (df['Owner'] != 4) | (df['Owner'] != 5) | (df['Owner'] != 6)] = 9

我使用的第二种方法是带有 if 语句的循环。但这对列表没有任何作用

for i in df['Owner']:
    if i == 1:
        pass
    elif i == 2:
        pass
    elif i == 3:
        pass
    elif i == 4:
        pass
    elif i == 5:
        pass
    elif i == 6:
        pass
    elif i == 0:
        pass
    else:
        i = 9

提前致谢

Answer 1

df['Owner'] = [10 if type(x) is not int else x for x in df['Owner']]

应该这样做。它读起来像 Python 中经常出现的简单英语。

Answer 2

你可以maptype函数，和int比较，然后用where保留整数，其余的用10代替：

df['Owner'] = df['Owner'].where(df['Owner'].map(type).eq(int), 10)

输出：

  Owner
0    10
1    10
2     2
3    10
4     5
5    10

使用过的输入：

from numpy import nan
df = pd.DataFrame({'Owner': ['Owned by MHPI Inc.', nan, 2, nan, 5, 'Owned by RHP Properties',]})