Python Pandas Counting the Occurrences of a Specific value

Question

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' 

Answer 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)') 

Answer 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