关于python：pandas：多列的to_numeric

pandas: to_numeric for multiple columns

我正在使用以下df：

1
2
3
4
5

c.sort_values('2005', ascending=False).head(3)
GeoName ComponentName IndustryId IndustryClassification Description 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014
37926 Alabama Real GDP by state 9 213 Support activities for mining 99 98 117 117 115 87 96 95 103 102 (NA)
37951 Alabama Real GDP by state 34 42 Wholesale trade 9898 10613 10952 11034 11075 9722 9765 9703 9600 9884 10199
37932 Alabama Real GDP by state 15 327 Nonmetallic mineral products manufacturing 980 968 940 1084 861 724 714 701 589 641 (NA)

我想在所有年份强制数字：

1	c['2014'] = pd.to_numeric(c['2014'], errors='coerce')

有没有一种简单的方法可以做到这一点，还是我必须全部输入？

更新：之后您无需转换您的值，您可以在阅读CSV时即时执行此操作：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21

In [165]: df=pd.read_csv(url, index_col=0, na_values=['(NA)']).fillna(0)

In [166]: df.dtypes
Out[166]:
GeoName object
ComponentName object
IndustryId int64
IndustryClassification object
Description object
2004 int64
2005 int64
2006 int64
2007 int64
2008 int64
2009 int64
2010 int64
2011 int64
2012 int64
2013 int64
2014 float64
dtype: object

如果需要将多个列转换为数字dtypes - 请使用以下技术：

样本来源DF：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19

In [271]: df
Out[271]:
id a b c d e f
0 id_3 AAA 6 3 5 8 1
1 id_9 3 7 5 7 3 BBB
2 id_7 4 2 3 5 4 2
3 id_0 7 3 5 7 9 4
4 id_0 2 4 6 4 0 2

In [272]: df.dtypes
Out[272]:
id object
a object
b int64
c int64
d int64
e int64
f object
dtype: object

将所选列转换为数字dtypes：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23

In [273]: cols = df.columns.drop('id')

In [274]: df[cols] = df[cols].apply(pd.to_numeric, errors='coerce')

In [275]: df
Out[275]:
id a b c d e f
0 id_3 NaN 6 3 5 8 1.0
1 id_9 3.0 7 5 7 3 NaN
2 id_7 4.0 2 3 5 4 2.0
3 id_0 7.0 3 5 7 9 4.0
4 id_0 2.0 4 6 4 0 2.0

In [276]: df.dtypes
Out[276]:
id object
a float64
b int64
c int64
d int64
e int64
f float64
dtype: object

PS如果要选择所有string(object)列，请使用以下简单技巧：

1	cols = df.columns[df.dtypes.eq('object')]

相关讨论

您可以使用：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23

print df.columns[5:]
Index([u'2004', u'2005', u'2006', u'2007', u'2008', u'2009', u'2010', u'2011',
u'2012', u'2013', u'2014'],
dtype='object')

for col in df.columns[5:]:
df[col] = pd.to_numeric(df[col], errors='coerce')

print df
GeoName ComponentName IndustryId IndustryClassification \
37926 Alabama Real GDP by state 9 213
37951 Alabama Real GDP by state 34 42
37932 Alabama Real GDP by state 15 327

Description 2004 2005 2006 2007 \
37926 Support activities for mining 99 98 117 117
37951 Wholesale trade 9898 10613 10952 11034
37932 Nonmetallic mineral products manufacturing 980 968 940 1084

2008 2009 2010 2011 2012 2013 2014
37926 115 87 96 95 103 102 NaN
37951 11075 9722 9765 9703 9600 9884 10199.0
37932 861 724 714 701 589 641 NaN

filter的另一个解决方案：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23

print df.filter(like='20')
2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014
37926 99 98 117 117 115 87 96 95 103 102 (NA)
37951 9898 10613 10952 11034 11075 9722 9765 9703 9600 9884 10199
37932 980 968 940 1084 861 724 714 701 589 641 (NA)

for col in df.filter(like='20').columns:
df[col] = pd.to_numeric(df[col], errors='coerce')
print df
GeoName ComponentName IndustryId IndustryClassification \
37926 Alabama Real GDP by state 9 213
37951 Alabama Real GDP by state 34 42
37932 Alabama Real GDP by state 15 327

Description 2004 2005 2006 2007 \
37926 Support activities for mining 99 98 117 117
37951 Wholesale trade 9898 10613 10952 11034
37932 Nonmetallic mineral products manufacturing 980 968 940 1084

2008 2009 2010 2011 2012 2013 2014
37926 115 87 96 95 103 102 NaN
37951 11075 9722 9765 9703 9600 9884 10199.0
37932 861 724 714 701 589 641 NaN