在每列中查找 DataFrame 中不同元素的计数

Question

I am trying to find the count of distinct values in each column using Pandas. This is what I did.

import pandas as pd
import numpy as np

# Generate data.
NROW = 10000
NCOL = 100
df = pd.DataFrame(np.random.randint(1, 100000, (NROW, NCOL)),
                  columns=['col' + x for x in np.arange(NCOL).astype(str)])

I need to count the number of distinct elements for each column, like this:

col0    9538
col1    9505
col2    9524

What would be the most efficient way to do this, as this method will be applied to files which have size greater than 1.5GB?

---

Based upon the answers, df.apply(lambda x: len(x.unique())) is the fastest ( notebook ).

%timeit df.apply(lambda x: len(x.unique())) 10 loops, best of 3: 49.5 ms per loop %timeit df.nunique() 10 loops, best of 3: 59.7 ms per loop %timeit df.apply(pd.Series.nunique) 10 loops, best of 3: 60.3 ms per loop %timeit df.T.apply(lambda x: x.nunique(), axis=1) 10 loops, best of 3: 60.5 ms per loop

Answer 1

As of pandas 0.20 we can use nunique directly on DataFrame s, ie:

df.nunique()
a    4
b    5
c    1
dtype: int64

Other legacy options:

You could do a transpose of the df and then using apply call nunique row-wise:

In [205]:
df = pd.DataFrame({'a':[0,1,1,2,3],'b':[1,2,3,4,5],'c':[1,1,1,1,1]})
df

Out[205]:
   a  b  c
0  0  1  1
1  1  2  1
2  1  3  1
3  2  4  1
4  3  5  1

In [206]:
df.T.apply(lambda x: x.nunique(), axis=1)

Out[206]:
a    4
b    5
c    1
dtype: int64

EDIT

As pointed out by @ajcr the transpose is unnecessary:

In [208]:
df.apply(pd.Series.nunique)

Out[208]:
a    4
b    5
c    1
dtype: int64

Answer 2

A Pandas.Series has a .value_counts() function that provides exactly what you want to. Check out the documentation for the function .

Answer 3

Already some great answers here :) but this one seems to be missing:

df.apply(lambda x: x.nunique())

As of pandas 0.20.0, DataFrame.nunique() is also available.

Answer 4

Recently, I have same issues of counting unique value of each column in DataFrame, and I found some other function that runs faster than the apply function:

#Select the way how you want to store the output, could be pd.DataFrame or Dict, I will use Dict to demonstrate:
col_uni_val={}
for i in df.columns:
    col_uni_val[i] = len(df[i].unique())

#Import pprint to display dic nicely:
import pprint
pprint.pprint(col_uni_val)

This works for me almost twice faster than df.apply(lambda x: len(x.unique()))

Answer 5

I found:

df.agg(['nunique']).T

much faster

Answer 6

df.apply(lambda x: len(x.unique()))

Answer 7

Need to segregate only the columns with more than 20 unique values for all the columns in pandas_python:

enter code here
col_with_morethan_20_unique_values_cat=[]
for col in data.columns:
    if data[col].dtype =='O':
        if len(data[col].unique()) >20:

        ....col_with_morethan_20_unique_values_cat.append(data[col].name)
        else:
            continue

print(col_with_morethan_20_unique_values_cat)
print('total number of columns with more than 20 number of unique value is',len(col_with_morethan_20_unique_values_cat))



 # The o/p will be as:
['CONTRACT NO', 'X2','X3',,,,,,,..]
total number of columns with more than 20 number of unique value is 25

Answer 8

Adding the example code for the answer given by @CaMaDuPe85

df = pd.DataFrame({'a':[0,1,1,2,3],'b':[1,2,3,4,5],'c':[1,1,1,1,1]})
df

# df
    a   b   c
0   0   1   1
1   1   2   1
2   1   3   1
3   2   4   1
4   3   5   1


for cs in df.columns:
    print(cs,df[cs].value_counts().count()) 
    # using value_counts in each column and count it 

# Output

a 4
b 5
c 1