Count appearances of a value until it changes to another value
Use:
df = df.groupby(df['values'].ne(df['values'].shift()).cumsum())['values'].value_counts()
Or:
df = df.groupby([df['values'].ne(df['values'].shift()).cumsum(), 'values']).size()
print (df)
values values
1 10 2
2 23 2
3 9 3
4 10 4
5 12 1
Name: values, dtype: int64
Last for remove first level:
df = df.reset_index(level=0, drop=True)
print (df)
values
10 2
23 2
9 3
10 4
12 1
dtype: int64
Explanation:
Compare original column by shift
ed with not equal ne
and then add cumsum
for helper Series
:
print (pd.concat([df['values'], a, b, c],
keys=('orig','shifted', 'not_equal', 'cumsum'), axis=1))
orig shifted not_equal cumsum
0 10 NaN True 1
1 10 10.0 False 1
2 23 10.0 True 2
3 23 23.0 False 2
4 9 23.0 True 3
5 9 9.0 False 3
6 9 9.0 False 3
7 10 9.0 True 4
8 10 10.0 False 4
9 10 10.0 False 4
10 10 10.0 False 4
11 12 10.0 True 5
You can keep track of where the changes in df['values']
occur, and groupby
the changes and also df['values']
(to keep them as index) computing the size
of each group
changes = df['values'].diff().ne(0).cumsum()
df.groupby([changes,'values']).size().reset_index(level=0, drop=True)
values
10 2
23 2
9 3
10 4
12 1
dtype: int64
itertools.groupby
from itertools import groupby
pd.Series(*zip(*[[len([*v]), k] for k, v in groupby(df['values'])]))
10 2
23 2
9 3
10 4
12 1
dtype: int64
It's a generator
def f(x):
count = 1
for this, that in zip(x, x[1:]):
if this == that:
count += 1
else:
yield count, this
count = 1
yield count, [*x][-1]
pd.Series(*zip(*f(df['values'])))
10 2
23 2
9 3
10 4
12 1
dtype: int64
Using crosstab
df['key']=df['values'].diff().ne(0).cumsum()
pd.crosstab(df['key'],df['values'])
Out[353]:
values 9 10 12 23
key
1 0 2 0 0
2 0 0 0 2
3 3 0 0 0
4 0 4 0 0
5 0 0 1 0
Slightly modify the result above
pd.crosstab(df['key'],df['values']).stack().loc[lambda x:x.ne(0)]
Out[355]:
key values
1 10 2
2 23 2
3 9 3
4 10 4
5 12 1
dtype: int64
Base on python
groupby
from itertools import groupby
[ (k,len(list(g))) for k,g in groupby(df['values'].tolist())]
Out[366]: [(10, 2), (23, 2), (9, 3), (10, 4), (12, 1)]