pandas: how to get the percentage for each row

Question

When I use pandas value_count method, I get the data below:

new_df['mark'].value_counts()

1   1349110
2   1606640
3   175629
4   790062
5   330978

How can I get the percentage for each row like this?

1   1349110 31.7%
2   1606640 37.8%
3   175629  4.1%
4   790062  18.6%
5   330978  7.8%

I need to divide each row by the sum of these data.

Asked By: liu gang

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Source

Answer 1

I think you need:

#if output is Series, convert it to DataFrame
df = df.rename('a').to_frame()

df['per'] = (df.a * 100 / df.a.sum()).round(1).astype(str) + '%'

print (df)
         a    per
1  1349110  31.7%
2  1606640  37.8%
3   175629   4.1%
4   790062  18.6%
5   330978   7.8%

Timings:

It seems faster is use sum as twice value_counts:

In [184]: %timeit (jez(s))
10 loops, best of 3: 38.9 ms per loop

In [185]: %timeit (pir(s))
10 loops, best of 3: 76 ms per loop

Code for timings:

np.random.seed([3,1415])
s = pd.Series(np.random.choice(list('ABCDEFGHIJ'), 1000, p=np.arange(1, 11) / 55.))
s = pd.concat([s]*1000)#.reset_index(drop=True)

def jez(s):
    df = s.value_counts()
    df = df.rename('a').to_frame()
    df['per'] = (df.a * 100 / df.a.sum()).round(1).astype(str) + '%'
    return df

def pir(s):
    return pd.DataFrame({'a':s.value_counts(), 
                         'per':s.value_counts(normalize=True).mul(100).round(1).astype(str) + '%'})

print (jez(s))
print (pir(s))

Answered By: jezrael

Answer 2

np.random.seed([3,1415])
s = pd.Series(np.random.choice(list('ABCDEFGHIJ'), 1000, p=np.arange(1, 11) / 55.))

s.value_counts()

I    176
J    167
H    136
F    128
G    111
E     85
D     83
C     52
B     38
A     24
dtype: int64

As percent

s.value_counts(normalize=True)

I    0.176
J    0.167
H    0.136
F    0.128
G    0.111
E    0.085
D    0.083
C    0.052
B    0.038
A    0.024
dtype: float64

counts = s.value_counts()
percent = counts / counts.sum()
fmt = '{:.1%}'.format
pd.DataFrame({'counts': counts, 'per': percent.map(fmt)})

   counts    per
I     176  17.6%
J     167  16.7%
H     136  13.6%
F     128  12.8%
G     111  11.1%
E      85   8.5%
D      83   8.3%
C      52   5.2%
B      38   3.8%
A      24   2.4%

Answered By: piRSquared

Answer 3

Here’s a more pythonic snippet than what is proposed above I think

def aspercent(column,decimals=2):
    assert decimals >= 0
    return (round(column*100,decimals).astype(str) + "%")

aspercent(df['mark'].value_counts(normalize=True),decimals=1)

This will output:

1   1349110 31.7%
2   1606640 37.8%
3   175629  4.1%
4   790062  18.6%
5   330978  7.8%

This also allows to adjust the number of decimals

Answered By: Delevoye Guillaume

Answer 4

Create two series, first one with absolute values and a second one with percentages, and concatenate them:

import pandas 

d = {'mark': ['pos', 'pos', 'pos', 'pos', 'pos', 
              'neg', 'neg', 'neg', 'neg', 
              'neutral', 'neutral' ]}
df = pd.DataFrame(d)

absolute = df['mark'].value_counts(normalize=False)
absolute.name = 'value'
percentage = df['mark'].value_counts(normalize=True)
percentage.name = '%'
percentage = (percentage*100).round(2)
pd.concat([absolute, percentage], axis=1)

Output:

       value    %
pos     5   45.45
neg     4   36.36
neutral 2   18.18

Answered By: ssoler

pandas: how to get the percentage for each row

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