Count unique values with pandas per groups [duplic

Question

This question already has an answer here:

• Pandas count(distinct) equivalent 4 answers

I need to count unique ID values in every domain I have data

ID, domain
123, 'vk.com'
123, 'vk.com'
123, 'twitter.com'
456, 'vk.com'
456, 'facebook.com'
456, 'vk.com'
456, 'google.com'
789, 'twitter.com'
789, 'vk.com'

I try df.groupby(['domain', 'ID']).count() But I want to get

domain, count
vk.com   3
twitter.com   2
facebook.com   1
google.com   1

Answer 1

Generally to count distinct values in single column, you can use Series.value_counts:

df.domain.value_counts()

#'vk.com'          5
#'twitter.com'     2
#'facebook.com'    1
#'google.com'      1
#Name: domain, dtype: int64

To see how many unique values in a column, use Series.nunique:

df.domain.nunique()
# 4

To get all these distinct values, you can use unique or drop_duplicates, the slight difference between the two functions is that unique return a numpy.array while drop_duplicates returns a pandas.Series:

df.domain.unique()
# array(["'vk.com'", "'twitter.com'", "'facebook.com'", "'google.com'"], dtype=object)

df.domain.drop_duplicates()
#0          'vk.com'
#2     'twitter.com'
#4    'facebook.com'
#6      'google.com'
#Name: domain, dtype: object

---

As for this specific problem, since you'd like to count distinct value with respect to another variable, besides groupby method provided by other answers here, you can also simply drop duplicates firstly and then do value_counts():

import pandas as pd
df.drop_duplicates().domain.value_counts()

# 'vk.com'          3
# 'twitter.com'     2
# 'facebook.com'    1
# 'google.com'      1
# Name: domain, dtype: int64

Answer 2

IIUC you want the number of different ID for every domain, then you can try this:

output = df.drop_duplicates()
output.groupby('domain').size()

output:

    domain
facebook.com    1
google.com      1
twitter.com     2
vk.com          3
dtype: int64

You could also use value_counts, which is slightly less efficient.But the best is Jezrael's answer using nunique:

%timeit df.drop_duplicates().groupby('domain').size()
1000 loops, best of 3: 939 µs per loop
%timeit df.drop_duplicates().domain.value_counts()
1000 loops, best of 3: 1.1 ms per loop
%timeit df.groupby('domain')['ID'].nunique()
1000 loops, best of 3: 440 µs per loop

Answer 3

df.domain.value_counts()

>>> df.domain.value_counts()

vk.com          5

twitter.com     2

google.com      1

facebook.com    1

Name: domain, dtype: int64

Answer 4

You need nunique:

df = df.groupby('domain')['ID'].nunique()

print (df)
domain
'facebook.com'    1
'google.com'      1
'twitter.com'     2
'vk.com'          3
Name: ID, dtype: int64

If you need to strip ' characters:

df = df.ID.groupby([df.domain.str.strip("'")]).nunique()
print (df)
domain
facebook.com    1
google.com      1
twitter.com     2
vk.com          3
Name: ID, dtype: int64

Or as Jon Clements commented:

df.groupby(df.domain.str.strip("'"))['ID'].nunique()

You can retain the column name like this:

df = df.groupby(by='domain', as_index=False).agg({'ID': pd.Series.nunique})
print(df)
    domain  ID
0       fb   1
1      ggl   1
2  twitter   2
3       vk   3

The difference is that nunique() returns a Series and agg() returns a DataFrame.