How to count unique records by two columns in pandas?

Question

I have dataframe in pandas:

In [10]: df
Out[10]:
    col_a    col_b  col_c  col_d
0  France    Paris      3      4
1      UK    Londo      4      5
2      US  Chicago      5      6
3      UK  Bristol      3      3
4      US    Paris      8      9
5      US   London     44      4
6      US  Chicago     12      4

I need to count unique cities. I can count unique states

In [11]: df['col_a'].nunique()
Out[11]: 3

and I can try to count unique cities

In [12]: df['col_b'].nunique()
Out[12]: 5

but it is wrong because US Paris and Paris in France are different cities. So now I'm doing in like this:

In [13]: df['col_a_b'] = df['col_a'] + ' - ' + df['col_b']

In [14]: df
Out[14]:
    col_a    col_b  col_c  col_d         col_a_b
0  France    Paris      3      4  France - Paris
1      UK    Londo      4      5      UK - Londo
2      US  Chicago      5      6    US - Chicago
3      UK  Bristol      3      3    UK - Bristol
4      US    Paris      8      9      US - Paris
5      US   London     44      4     US - London
6      US  Chicago     12      4    US - Chicago

In [15]: df['col_a_b'].nunique()
Out[15]: 6

Maybe there is a better way? Without creating an additional column.

Answer 1

import pandas as pd
data = {'field1':[1,4,1,68,9],'field2':[1,1,4,5,9]}
df = pd.DataFrame(data)
results = df.groupby('field1')['field2'].nunique()
results

Output:

field1
1     2
4     1
9     1
68    1
Name: field2, dtype: int64

Answer 2

try this, I'm basically subtracting the number of duplicate groups from the number of rows in df. This is assuming we are grouping all the categories in the df

df.shape[0] - df[['col_a','col_b']].duplicated().sum()

774 µs ± 603 ns per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Answer 3

You can select col_a and col_b, drop the duplicates, then check the shape/len of the result data frame:

df[['col_a', 'col_b']].drop_duplicates().shape[0]
# 6

len(df[['col_a', 'col_b']].drop_duplicates())
# 6

---

Because groupby ignore NaNs, and may unnecessarily invoke a sorting process, choose accordingly which method to use if you have NaNs in the columns:

Consider a data frame as following:

df = pd.DataFrame({
    'col_a': [1,2,2,pd.np.nan,1,4],
    'col_b': [2,2,3,pd.np.nan,2,pd.np.nan]
})

print(df)

#   col_a  col_b
#0    1.0    2.0
#1    2.0    2.0
#2    2.0    3.0
#3    NaN    NaN
#4    1.0    2.0
#5    4.0    NaN

Timing:

df = pd.concat([df] * 1000)

%timeit df.groupby(['col_a', 'col_b']).ngroups
# 1000 loops, best of 3: 625 µs per loop

%timeit len(df[['col_a', 'col_b']].drop_duplicates())
# 1000 loops, best of 3: 1.02 ms per loop

%timeit df[['col_a', 'col_b']].drop_duplicates().shape[0]
# 1000 loops, best of 3: 1.01 ms per loop    

%timeit len(set(zip(df['col_a'],df['col_b'])))
# 10 loops, best of 3: 56 ms per loop

%timeit len(df.groupby(['col_a', 'col_b']))
# 1 loop, best of 3: 260 ms per loop

Result:

df.groupby(['col_a', 'col_b']).ngroups
# 3

len(df[['col_a', 'col_b']].drop_duplicates())
# 5

df[['col_a', 'col_b']].drop_duplicates().shape[0]
# 5

len(set(zip(df['col_a'],df['col_b'])))
# 2003

len(df.groupby(['col_a', 'col_b']))
# 2003

So the difference:

Option 1:

df.groupby(['col_a', 'col_b']).ngroups

is fast, and it excludes rows that contain NaNs.

Option 2 & 3:

len(df[['col_a', 'col_b']].drop_duplicates())
df[['col_a', 'col_b']].drop_duplicates().shape[0]

Reasonably fast, it considers NaNs as a unique value.

Option 4 & 5:

len(set(zip(df['col_a'],df['col_b']))) 
len(df.groupby(['col_a', 'col_b'])) 

slow, and it is following the logic that numpy.nan == numpy.nan is False, so different (nan, nan) rows are considered different.

Answer 4

By using ngroups

df.groupby(['col_a', 'col_b']).ngroups
Out[101]: 6

Or using set

len(set(zip(df['col_a'],df['col_b'])))
Out[106]: 6

Answer 5

In [105]: len(df.groupby(['col_a', 'col_b']))
Out[105]: 6