Pandas cumulative count [duplicate]

Answer 1

You can just compare the column against the value of interest and call cumsum:

In [12]:
df['obj1_count'] = (df['object'] == 'obj1').cumsum()
df['obj2_count'] = (df['object'] == 'obj2').cumsum()
df

Out[12]:
      time object  obj1_count  obj2_count
idx                                      
0    04:10   obj1           1           0
1    04:10   obj1           2           0
2    04:11   obj1           3           0
3    04:12   obj2           3           1
4    04:12   obj2           3           2
5    04:12   obj1           4           2
6    04:13   obj2           4           3

Here the comparison will produce a boolean series:

In [13]:
df['object'] == 'obj1'

Out[13]:
idx
0     True
1     True
2     True
3    False
4    False
5     True
6    False
Name: object, dtype: bool

when you call cumsum on the above the True values are converted to 1 and False to 0 and are summed cumulatively

Answer 2

You can generalize this process by getting the cumsum of pd.get_dummies. This should work for an arbitrary number of objects you want to count, without needing to specify each one individually:

# Get the cumulative counts.
counts = pd.get_dummies(df['object']).cumsum()

# Rename the count columns as appropriate.
counts = counts.rename(columns=lambda col: col+'_count')

# Join the counts to the original df.
df = df.join(counts)

The resulting output:

    time object  obj1_count  obj2_count
0  04:10   obj1           1           0
1  04:10   obj1           2           0
2  04:11   obj1           3           0
3  04:12   obj2           3           1
4  04:12   obj2           3           2
5  04:12   obj1           4           2
6  04:13   obj2           4           3

You can omit the rename step if it's acceptable to use count as a prefix instead of a suffix, i.e. 'count_obj1' instead of 'obj1_count'. Simply use the prefix parameter of pd.get_dummies:

 counts = pd.get_dummies(df['object'], prefix='count').cumsum()

Answer 3

Here's a way using numpy

u, iv = np.unique(
    df.object.values,
    return_inverse=True
)

objcount = pd.DataFrame(
    (iv[:, None] == np.arange(len(u))).cumsum(0),
    df.index, u
)
pd.concat([df, objcount], axis=1)

Answer 4

There is a special function for such operation: cumcount

>>> df = pd.DataFrame([['a'], ['a'], ['a'], ['b'], ['b'], ['a']], columns=['A'])
>>> df
   A
0  a
1  a
2  a
3  b
4  b
5  a
>>> df.groupby('A').cumcount()
0    0
1    1
2    2
3    0
4    1
5    3
dtype: int64
>>> df.groupby('A').cumcount(ascending=False)
0    3
1    2
2    1
3    1
4    0
5    0
 dtype: int64