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.
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
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 NaN
s, and may unnecessarily invoke a sorting process, choose accordingly which method to use if you have NaN
s 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 NaN
s.
Option 2 & 3:
len(df[['col_a', 'col_b']].drop_duplicates())
df[['col_a', 'col_b']].drop_duplicates().shape[0]
Reasonably fast, it considers NaN
s 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.
In [105]: len(df.groupby(['col_a', 'col_b']))
Out[105]: 6
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)
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
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