function outputs the number of all unique values in a column, for example

 apple   3
 orange  2
 banana  1

I want to search the total number of (value = 'apple') only, which function can replace value_counts()?

  • $\begingroup$ Hi! It would have helped me if you would provide the declaration of your data structure. I think your series is actually a column in a dataframe, but you'll manage. $\endgroup$ Commented Aug 13, 2019 at 11:16

2 Answers 2


You have plenty of ways to do it. You won't see a big difference in performance. My suggestion is to use whatever feels more convenient for you or your team.

import pandas as pd
import numpy as np

#Let's create a dataframe with 10 million integers from 0 to 100
df = pd.DataFrame(np.random.randint(0,100,size=(10000000, 1)), columns=list('A'))

#And now count the value 5 with 4 different ways

%timeit df[df.A == 5].shape[0]
10 loops, best of 3: 25.4 ms per loop

%timeit len(df[df.A == 5])
10 loops, best of 3: 25.4 ms per loop

%timeit len(df[df.A == 5].index)
10 loops, best of 3: 25.6 ms per loop

%timeit df['A'].value_counts()[5]
10 loops, best of 3: 149 ms per loop

As you can see, only the last one takes more time to run.

EDIT: Addition to your comment, you could try this

df = data.groupby('a_1').get_group(a_2)['suffix']
len(df[df.suffix == 'a_3'])
  • $\begingroup$ i changed my code to data.groupby('a_1').get_group(a_2)['suffix'].value_counts()['a_3'], it works. i tried to changed to other 3 faster functions. how? $\endgroup$
    – TJCLK
    Commented Aug 14, 2019 at 2:22
  • 1
    $\begingroup$ @LiDong please check my edit $\endgroup$
    – Tasos
    Commented Aug 14, 2019 at 5:47

You can filter a Series, then apply the value_count.


fruits[fruits == "apple"].value_counts()
  • 1
    $\begingroup$ Instead of using value_counts() you can just use len() function on the filtered result $\endgroup$
    – vico
    Commented Aug 13, 2019 at 11:47
  • $\begingroup$ Yes you are right. $\endgroup$ Commented Aug 13, 2019 at 13:22

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