问题
I have the following DataFrame:
df = pd.DataFrame([10, 10, 23, 23, 9, 9, 9, 10, 10, 10, 10, 12], columns=['values'])
I want to calculate the frequency of each value, but not an overall count - the count of each value until it changes to another value.
I tried:
df['values'].value_counts()
but it gives me
10 6
9 3
23 2
12 1
The desired output is
10:2
23:2
9:3
10:4
12:1
How can I do this?
回答1:
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 shifted 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
回答2:
You can keep track of where the changes in df['values'] occur:
changes = df['values'].diff().ne(0).cumsum()
print(changes)
0 1
1 1
2 2
3 2
4 3
5 3
6 3
7 4
8 4
9 4
10 4
11 5
And groupby the changes and also df['values'] (to keep them as index) computing the size of each group
df.groupby([changes,'values']).size().reset_index(level=0, drop=True)
values
10 2
23 2
9 3
10 4
12 1
dtype: int64
回答3:
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
回答4:
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)]
回答5:
This is far from the most time/memory efficient method that in this thread but here's an iterative approach that is pretty straightforward. Please feel encouraged to suggest improvements on this method.
import pandas as pd
df = pd.DataFrame([10, 10, 23, 23, 9, 9, 9, 10, 10, 10, 10, 12], columns=['values'])
dict_count = {}
for v in df['values'].unique():
dict_count[v] = 0
curr_val = df.iloc[0]['values']
count = 1
for i in range(1, len(df)):
if df.iloc[i]['values'] == curr_val:
count += 1
else:
if count > dict_count[curr_val]:
dict_count[curr_val] = count
curr_val = df.iloc[i]['values']
count = 1
if count > dict_count[curr_val]:
dict_count[curr_val] = count
df_count = pd.DataFrame(dict_count, index=[0])
print(df_count)
回答6:
The function groupby in itertools can help you, for str:
>>> string = 'aabbaacc'
>>> for char, freq in groupby('aabbaacc'):
>>> print(char, len(list(freq)), sep=':', end='\n')
[out]:
a:2
b:2
a:2
c:2
This function also works for list:
>>> df = pd.DataFrame([10, 10, 23, 23, 9, 9, 9, 10, 10, 10, 10, 12], columns=['values'])
>>> for char, freq in groupby(df['values'].tolist()):
>>> print(char, len(list(freq)), sep=':', end='\n')
[out]:
10:2
23:2
9:3
10:4
12:1
Note: for df, you always use this way like df['values'] to take 'values' column, because DataFrame have a attribute values
来源：https://stackoverflow.com/questions/53542668/count-appearances-of-a-value-until-it-changes-to-another-value