How to groupby consecutive values in pandas DataFrame

Question

I have a column in a DataFrame with values:

[1, 1, -1, 1, -1, -1]

How can I group them like this?

[1,1] [-1] [1] [-1, -1]

Solution 1

You can use groupby by custom Series:

df = pd.DataFrame({'a': [1, 1, -1, 1, -1, -1]})
print (df)
   a
0  1
1  1
2 -1
3  1
4 -1
5 -1

print ((df.a != df.a.shift()).cumsum())
0    1
1    1
2    2
3    3
4    4
5    4
Name: a, dtype: int32

for i, g in df.groupby([(df.a != df.a.shift()).cumsum()]):
    print (i)
    print (g)
    print (g.a.tolist())

   a
0  1
1  1
[1, 1]
2
   a
2 -1
[-1]
3
   a
3  1
[1]
4
   a
4 -1
5 -1
[-1, -1]

Solution 2

Using groupby from itertools data from Jez

from itertools import groupby
[ list(group) for key, group in groupby(df.a.values.tolist())]
Out[361]: [[1, 1], [-1], [1], [-1, -1]]

Solution 3

Series.diff is another way to mark the group boundaries (a!=a.shift means a.diff!=0):

consecutives = df['a'].diff().ne(0).cumsum()

# 0    1
# 1    1
# 2    2
# 3    3
# 4    4
# 5    4
# Name: a, dtype: int64

And to turn these groups into a Series of lists (see the other answers for a list of lists), aggregate with groupby.agg or groupby.apply:

df['a'].groupby(consecutives).agg(list)

# a
# 1      [1, 1]
# 2        [-1]
# 3         [1]
# 4    [-1, -1]
# Name: a, dtype: object

Comments

[Eran H.]

In case you want to use this solution to .groupby() consecutive dates with 1 hour difference, change the condition to df['date'].diff() != pd.Timedelta('1 hour')

[XoXo]

this answer is more explicit than the accepted cumsum() solution

[XoXo]

from the document: The operation of groupby() is similar to the uniq filter in Unix. It generates a break or new group every time the value of the key function changes

[XoXo]

github.com/pandas-dev/pandas/issues/5494 asks for the same behaviour with the itertools.groupby(), but it's Contributions Welcome, No action on 6 Jul 2018

[smci]

Instead of ==, there's actually a vectorized .ne() function: df.a.ne(df.a.shift())

[Rich Andrews]

While this is a literal answer to the question, it loses the oft-needed labeling of the group of consecutive values.