Python Pandas Counting the Occurrences of a Specific value

I am trying to find the number of times a certain value appears in one column.

I have made the dataframe with data = pd.DataFrame.from_csv('data/DataSet2.csv')

and now I want to find the number of times something appears in a column. How is this done?

I thought it was the below, where I am looking in the education column and counting the number of time ? occurs.

The code below shows that I am trying to find the number of times 9th appears and the error is what I am getting when I run the code

Code

missing2 = df.education.value_counts()['9th']
print(missing2)

Error

KeyError: '9th'

Solution 1:

You can create subset of data with your condition and then use shape or len:

print df
  col1 education
0    a       9th
1    b       9th
2    c       8th

print df.education == '9th'
0     True
1     True
2    False
Name: education, dtype: bool

print df[df.education == '9th']
  col1 education
0    a       9th
1    b       9th

print df[df.education == '9th'].shape[0]
2
print len(df[df['education'] == '9th'])
2

Performance is interesting, the fastest solution is compare numpy array and sum:

Code:

import perfplot, string
np.random.seed(123)


def shape(df):
    return df[df.education == 'a'].shape[0]

def len_df(df):
    return len(df[df['education'] == 'a'])

def query_count(df):
    return df.query('education == "a"').education.count()

def sum_mask(df):
    return (df.education == 'a').sum()

def sum_mask_numpy(df):
    return (df.education.values == 'a').sum()

def make_df(n):
    L = list(string.ascii_letters)
    df = pd.DataFrame(np.random.choice(L, size=n), columns=['education'])
    return df

perfplot.show(
    setup=make_df,
    kernels=[shape, len_df, query_count, sum_mask, sum_mask_numpy],
    n_range=[2**k for k in range(2, 25)],
    logx=True,
    logy=True,
    equality_check=False, 
    xlabel='len(df)')

Solution 2:

Couple of ways using count or sum

In [338]: df
Out[338]:
  col1 education
0    a       9th
1    b       9th
2    c       8th

In [335]: df.loc[df.education == '9th', 'education'].count()
Out[335]: 2

In [336]: (df.education == '9th').sum()
Out[336]: 2

In [337]: df.query('education == "9th"').education.count()
Out[337]: 2