Count frequency of values in pandas DataFrame column

PythonDjangoPandasDataframe

Python Problem Overview


I want to count number of times each values is appearing in dataframe.

Here is my dataframe - df:

    status
1     N
2     N
3     C
4     N
5     S
6     N
7     N
8     S
9     N
10    N
11    N
12    S
13    N
14    C
15    N
16    N
17    N
18    N
19    S
20    N

I want to dictionary of counts:

ex. counts = {N: 14, C:2, S:4}

I have tried df['status']['N'] but it gives keyError and also df['status'].value_counts but no use.

Python Solutions


Solution 1 - Python

You can use value_counts and to_dict:

print df['status'].value_counts()
N    14
S     4
C     2
Name: status, dtype: int64

counts = df['status'].value_counts().to_dict()
print counts
{'S': 4, 'C': 2, 'N': 14}

Solution 2 - Python

An alternative one liner using underdog Counter:

In [3]: from collections import Counter

In [4]: dict(Counter(df.status))
Out[4]: {'C': 2, 'N': 14, 'S': 4}

Solution 3 - Python

You can try this way.

df.stack().value_counts().to_dict()

Solution 4 - Python

Can you convert df into a list?

If so:

a = ['a', 'a', 'a', 'b', 'b', 'c']
c = dict()
for i in set(a):
    c[i] = a.count(i)

Using a dict comprehension:

c = {i: a.count(i) for i in set(a)}

Solution 5 - Python

See my response in this thread for a Pandas DataFrame output,

https://stackoverflow.com/questions/22391433/count-the-frequency-that-a-value-occurs-in-a-dataframe-column/55952449#55952449

For dictionary output, you can modify as follows:

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

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionKishan MehtaView Question on Stackoverflow
Solution 1 - PythonjezraelView Answer on Stackoverflow
Solution 2 - PythonColonel BeauvelView Answer on Stackoverflow
Solution 3 - Pythonsu79eu7kView Answer on Stackoverflow
Solution 4 - PythonChuckView Answer on Stackoverflow
Solution 5 - PythondjogunsView Answer on Stackoverflow