Python Pandas: remove entries based on the number of occurrences

Question

I\'m trying to remove entries from a data frame which occur less than 100 times. The data frame data looks like this:

pid   tag
1     23    
1          


        

Answer 1

Here are some run times for a couple of the solutions posted here, along with one that was not (using value_counts()) that is much faster than the other solutions:

Create the data:

import pandas as pd
import numpy as np

# Generate some 'users'
np.random.seed(42)
df = pd.DataFrame({'uid': np.random.randint(0, 500, 500)})

# Prove that some entries are 1
print "{:,} users only occur once in dataset".format(sum(df.uid.value_counts() == 1))

Output:

171 users only occur once in dataset

Time a few different ways of removing users with only one entry. These were run in separate cells in a Jupyter Notebook:

%%timeit
df.groupby(by='uid').filter(lambda x: len(x) > 1)

%%timeit
df[df.groupby('uid').uid.transform(len) > 1]

%%timeit
vc = df.uid.value_counts()
df[df.uid.isin(vc.index[vc.values > 1])].uid.value_counts()

These gave the following outputs:

10 loops, best of 3: 46.2 ms per loop
10 loops, best of 3: 30.1 ms per loop
1000 loops, best of 3: 1.27 ms per loop

Answer 2

Edit: Thanks to @WesMcKinney for showing this much more direct way:

data[data.groupby('tag').pid.transform(len) > 1]

---

import pandas
import numpy as np
data = pandas.DataFrame(
    {'pid' : [1,1,1,2,2,3,3,3],
     'tag' : [23,45,62,24,45,34,25,62],
     })

bytag = data.groupby('tag').aggregate(np.count_nonzero)
tags = bytag[bytag.pid >= 2].index
print(data[data['tag'].isin(tags)])

yields

   pid  tag
1    1   45
2    1   62
4    2   45
7    3   62

Answer 3

New in 0.12, groupby objects have a filter method, allowing you to do these types of operations:

In [11]: g = data.groupby('tag')

In [12]: g.filter(lambda x: len(x) > 1)  # pandas 0.13.1
Out[12]:
   pid  tag
1    1   45
2    1   62
4    2   45
7    3   62

The function (the first argument of filter) is applied to each group (subframe), and the results include elements of the original DataFrame belonging to groups which evaluated to True.

Note: in 0.12 the ordering is different than in the original DataFrame, this was fixed in 0.13+:

In [21]: g.filter(lambda x: len(x) > 1)  # pandas 0.12
Out[21]: 
   pid  tag
1    1   45
4    2   45
2    1   62
7    3   62

Answer 4

df = pd.DataFrame([(1, 2), (1, 3), (1, 4), (2, 1),(2,2,)], columns=['col1', 'col2'])

In [36]: df
Out[36]: 
   col1  col2
0     1     2
1     1     3
2     1     4
3     2     1
4     2     2

gp = df.groupby('col1').aggregate(np.count_nonzero)

In [38]: gp
Out[38]: 
      col2
col1      
1        3
2        2

lets get where the count > 2

tf = gp[gp.col2 > 2].reset_index()
df[df.col1 == tf.col1]

Out[41]: 
   col1  col2
0     1     2
1     1     3
2     1     4