Python 熊猫：多列的 to_numeric
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pandas: to_numeric for multiple columns
提问by Michael Perdue
I'm working with the following df:
我正在使用以下df：
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)
I want to force numeric on all of the years:
我想在所有年份都强制使用数字：
c['2014'] = pd.to_numeric(c['2014'], errors='coerce')
is there an easy way to do this or do I have to type them all out?
有没有一种简单的方法可以做到这一点，还是我必须将它们全部输入？
回答by MaxU
UPDATE:you don't need to convert your values afterwards, you can do it on-the-flywhen reading your CSV:
更新：之后您不需要转换您的值，您可以在读取 CSV 时即时进行：
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
If you need to convert multiple columns to numeric dtypes - use the following technique:
如果您需要将多列转换为数字 dtypes - 使用以下技术：
Sample source DF:
示例源 DF：
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
Converting selected columns to numeric dtypes:
将选定的列转换为数字 dtypes：
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 if you want to select allstring(object) columns use the following simple trick:
PS，如果您想选择所有string( object) 列，请使用以下简单技巧：
cols = df.columns[df.dtypes.eq('object')]
回答by muon
another way is using apply, one liner:
另一种方法是使用apply，一个班轮：
cols = ['col1', 'col2', 'col3']
data[cols] = data[cols].apply(pd.to_numeric, errors='coerce', axis=1)
回答by jezrael
You can use:
您可以使用：
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
Another solution with filter:
另一个解决方案filter：
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
回答by Ali Siddiq
df[cols] = pd.to_numeric(df[cols].stack(), errors='coerce').unstack()
回答by Jeremy
If you are looking for a range of columns, you can try this:
如果您正在寻找一系列列，您可以试试这个：
df.iloc[7:] = df.iloc[7:].astype(float)
The examples above will convert type to be float, for all the columns begin with the 7th to the end. You of course can use different type or different range.
上面的示例将类型转换为浮点数，因为所有列都从第 7 行开始到末尾。您当然可以使用不同的类型或不同的范围。
I think this is useful when you have a big range of columns to convert and a lot of rows. It doesn't make you go over each row by yourself - I believe numpy do it more efficiently.
我认为当您要转换的列范围很大且行很多时，这很有用。它不会让你自己遍历每一行 - 我相信 numpy 会更有效地完成它。
This is useful only if you know that all the required columns contain numbers only - it will not change "bad values" (like string) to be NaN for you.
仅当您知道所有必需的列仅包含数字时，这才有用 - 它不会将“坏值”（如字符串）更改为 NaN。
