统计一个值在 dataframe 列中出现的频率

Question

I have a dataset

category
cat a
cat b
cat a

I'd like to be able to return something like (showing unique values and frequency)

category   freq 
cat a       2
cat b       1

Answer 1

Use groupby and count :

In [37]:
df = pd.DataFrame({'a':list('abssbab')})
df.groupby('a').count()

Out[37]:

   a
a   
a  2
b  3
s  2

[3 rows x 1 columns]

See the online docs: https://pandas.pydata.org/pandas-docs/stable/user_guide/groupby.html

Also value_counts() as @DSM has commented, many ways to skin a cat here

In [38]:
df['a'].value_counts()

Out[38]:

b    3
a    2
s    2
dtype: int64

If you wanted to add frequency back to the original dataframe use transform to return an aligned index:

In [41]:
df['freq'] = df.groupby('a')['a'].transform('count')
df

Out[41]:

   a freq
0  a    2
1  b    3
2  s    2
3  s    2
4  b    3
5  a    2
6  b    3

[7 rows x 2 columns]

Answer 2

If you want to apply to all columns you can use:

df.apply(pd.value_counts)

This will apply a column based aggregation function (in this case value_counts) to each of the columns.

Answer 3

df.category.value_counts()

This short little line of code will give you the output you want.

If your column name has spaces you can use

df['category'].value_counts()

Answer 4

df.apply(pd.value_counts).fillna(0)

value_counts - Returns object containing counts of unique values

apply - count frequency in every column. If you set axis=1 , you get frequency in every row

fillna(0) - make output more fancy. Changed NaN to 0

Answer 5

In 0.18.1 groupby together with count does not give the frequency of unique values:

>>> df
   a
0  a
1  b
2  s
3  s
4  b
5  a
6  b

>>> df.groupby('a').count()
Empty DataFrame
Columns: []
Index: [a, b, s]

However, the unique values and their frequencies are easily determined using size :

>>> df.groupby('a').size()
a
a    2
b    3
s    2

With df.a.value_counts() sorted values (in descending order, ie largest value first) are returned by default.

Answer 6

Using list comprehension and value_counts for multiple columns in a df

[my_series[c].value_counts() for c in list(my_series.select_dtypes(include=['O']).columns)]

https://stackoverflow.com/a/28192263/786326

Answer 7

If your DataFrame has values with the same type, you can also set return_counts=True in numpy.unique() .

index, counts = np.unique(df.values,return_counts=True)

np.bincount() could be faster if your values are integers.

Answer 8

Without any libraries, you could do this instead:

def to_frequency_table(data):
    frequencytable = {}
    for key in data:
        if key in frequencytable:
            frequencytable[key] += 1
        else:
            frequencytable[key] = 1
    return frequencytable

Example:

to_frequency_table([1,1,1,1,2,3,4,4])
>>> {1: 4, 2: 1, 3: 1, 4: 2}

Answer 9

You can also do this with pandas by broadcasting your columns as categories first, eg dtype="category" eg

cats = ['client', 'hotel', 'currency', 'ota', 'user_country']

df[cats] = df[cats].astype('category')

and then calling describe :

df[cats].describe()

This will give you a nice table of value counts and a bit more :):

    client  hotel   currency    ota user_country
count   852845  852845  852845  852845  852845
unique  2554    17477   132 14  219
top 2198    13202   USD Hades   US
freq    102562  8847    516500  242734  340992

Answer 10

@metatoaster has already pointed this out. Go for Counter . It's blazing fast.

import pandas as pd
from collections import Counter
import timeit
import numpy as np

df = pd.DataFrame(np.random.randint(1, 10000, (100, 2)), columns=["NumA", "NumB"])

Timers

%timeit -n 10000 df['NumA'].value_counts()
# 10000 loops, best of 3: 715 µs per loop

%timeit -n 10000 df['NumA'].value_counts().to_dict()
# 10000 loops, best of 3: 796 µs per loop

%timeit -n 10000 Counter(df['NumA'])
# 10000 loops, best of 3: 74 µs per loop

%timeit -n 10000 df.groupby(['NumA']).count()
# 10000 loops, best of 3: 1.29 ms per loop

Cheers!

Answer 11

I believe this should work fine for any DataFrame columns list.

def column_list(x):
    column_list_df = []
    for col_name in x.columns:
        y = col_name, len(x[col_name].unique())
        column_list_df.append(y)
return pd.DataFrame(column_list_df)

column_list_df.rename(columns={0: "Feature", 1: "Value_count"})

The function "column_list" checks the columns names and then checks the uniqueness of each column values.

Answer 12

The following code creates frequency table for the various values in a column called "Total_score" in a dataframe called "smaller_dat1", and then returns the number of times the value "300" appears in the column.

valuec = smaller_dat1.Total_score.value_counts()
valuec.loc[300]

Answer 13

n_values = data.income.value_counts()

First unique value count

n_at_most_50k = n_values[0]

Second unique value count

n_greater_50k = n_values[1]

n_values

Output:

<=50K    34014
>50K     11208

Name: income, dtype: int64

Output:

n_greater_50k,n_at_most_50k:-
(11208, 34014)

Answer 14

Use this code:

import numpy as np
np.unique(df['a'],return_counts=True)

Answer 15

your data:

|category|
cat a
cat b
cat a

solution:

 df['freq'] = df.groupby('category')['category'].transform('count')
 df =  df.drop_duplicates()

Answer 16

As everyone said, the faster solution is to do:

df.column_to_analyze.value_counts()

But if you want to use the output in your dataframe, with this schema:

df input:

category
cat a
cat b
cat a

df output: 

category   counts
cat a        2
cat b        1 
cat a        2

you can do this:

df['counts'] = df.category.map(df.category.value_counts())
df