Pandas, Get count of a single value in a Column of a Dataframe

If you take the value_counts return, you can query it for multiple values:

import pandas as pd

a = pd.Series([1, 1, 1, 1, 2, 2])
counts = a.value_counts()
>>> counts[1], counts[2]
(4, 2)

However, to count only a single item, it would be faster to use

import numpy as np
np.sum(a == 1)

Get the total count:

column = df['specific_column']

column.count() 

Get the specific value total count:

column.loc[specific_column > 0].count() 

You do not need to add quotes ('') to indicate specific_column.

You can try value_counts:

df = df['col'].value_counts().reset_index()
df.columns = ['col', 'count']
print df
   col  count
0    1      5
1    2      3

EDIT:

print (df['col'] == 1).sum()
5

Or:

def somecalulation(x):
    return (df['col'] == x).sum()

print somecalulation(1)
5
print somecalulation(2)
3

Or:

ser = df['col'].value_counts()

def somecalulation(s, x):
    return s[x]

print somecalulation(ser, 1)
5
print somecalulation(ser, 2)
3

EDIT2:

If you need something really fast, use numpy.in1d:

import pandas as pd
import numpy as np

a = pd.Series([1, 1, 1, 1, 2, 2])

#for testing len(a) = 6000
a = pd.concat([a]*1000).reset_index(drop=True)

print np.in1d(a,1).sum()
4000
print (a == 1).sum()
4000
print np.sum(a==1)
4000

Timings:

len(a)=6:

In [131]: %timeit np.in1d(a,1).sum()
The slowest run took 9.17 times longer than the fastest. This could mean that an intermediate result is being cached 
10000 loops, best of 3: 29.9 µs per loop

In [132]: %timeit np.sum(a == 1)
10000 loops, best of 3: 196 µs per loop

In [133]: %timeit (a == 1).sum()
1000 loops, best of 3: 180 µs per loop

len(a)=6000:

In [135]: %timeit np.in1d(a,1).sum()
The slowest run took 7.29 times longer than the fastest. This could mean that an intermediate result is being cached 
10000 loops, best of 3: 48.5 µs per loop

In [136]: %timeit np.sum(a == 1)
The slowest run took 5.23 times longer than the fastest. This could mean that an intermediate result is being cached 
1000 loops, best of 3: 273 µs per loop

In [137]: %timeit (a == 1).sum()
1000 loops, best of 3: 271 µs per loop