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is there a simpler way to group and count with python?

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0

I am grouping and counting a set of data.

df = pd.DataFrame('key': ['A', 'B', 'A'],
 'data': np.ones(3,))
df.groupby('key').count()

outputs

 data
key 
A 2
B 1

The piece of code above works though, I wonder if there is a simpler one.

'data': np.ones(3,) seems to be a placeholder and indispensable.

pd.DataFrame(['A', 'B', 'A']).groupby(0).count()

outputs

A
B

My question is, is there a simpler way to do this, produce the count of 'A' and 'B' respectively, without something like 'data': np.ones(3,) ?

It doesn't have to be a pandas method, numpy or python native function are also appreciated.

python pandas numpy

share|improve this question

edited Mar 7 at 2:07

Ced

456418

asked Mar 7 at 0:22

shiqangpanshiqangpan

123

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  df.key.value_counts()?
  
  – coldspeed
  Mar 7 at 0:23
  
  

  
  
  
  

add a comment | 

0

I am grouping and counting a set of data.

df = pd.DataFrame('key': ['A', 'B', 'A'],
 'data': np.ones(3,))
df.groupby('key').count()

outputs

 data
key 
A 2
B 1

The piece of code above works though, I wonder if there is a simpler one.

'data': np.ones(3,) seems to be a placeholder and indispensable.

pd.DataFrame(['A', 'B', 'A']).groupby(0).count()

outputs

A
B

My question is, is there a simpler way to do this, produce the count of 'A' and 'B' respectively, without something like 'data': np.ones(3,) ?

It doesn't have to be a pandas method, numpy or python native function are also appreciated.

python pandas numpy

share|improve this question

edited Mar 7 at 2:07

Ced

456418

asked Mar 7 at 0:22

shiqangpanshiqangpan

123

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  df.key.value_counts()?
  
  – coldspeed
  Mar 7 at 0:23
  
  

  
  
  
  

add a comment | 

I am grouping and counting a set of data.

df = pd.DataFrame('key': ['A', 'B', 'A'],
 'data': np.ones(3,))
df.groupby('key').count()

outputs

 data
key 
A 2
B 1

The piece of code above works though, I wonder if there is a simpler one.

'data': np.ones(3,) seems to be a placeholder and indispensable.

pd.DataFrame(['A', 'B', 'A']).groupby(0).count()

outputs

A
B

My question is, is there a simpler way to do this, produce the count of 'A' and 'B' respectively, without something like 'data': np.ones(3,) ?

It doesn't have to be a pandas method, numpy or python native function are also appreciated.

python pandas numpy

share|improve this question

edited Mar 7 at 2:07

Ced

456418

asked Mar 7 at 0:22

shiqangpanshiqangpan

123

df = pd.DataFrame('key': ['A', 'B', 'A'],
 'data': np.ones(3,))
df.groupby('key').count()

 data
key 
A 2
B 1

pd.DataFrame(['A', 'B', 'A']).groupby(0).count()

A
B

share|improve this question

edited Mar 7 at 2:07

Ced

456418

asked Mar 7 at 0:22

shiqangpanshiqangpan

123

share|improve this question

edited Mar 7 at 2:07

Ced

456418

asked Mar 7 at 0:22

shiqangpanshiqangpan

123

edited Mar 7 at 2:07

Ced

456418

edited Mar 7 at 2:07

Ced

456418

asked Mar 7 at 0:22

shiqangpanshiqangpan

123

asked Mar 7 at 0:22

shiqangpanshiqangpan

123

3 Answers
3

active

oldest

votes

0

Use a Series instead.

>>> import pandas as pd
>>> 
>>> data = ['A', 'A', 'A', 'B', 'C', 'C', 'D', 'D', 'D', 'D', 'D']
>>> 
>>> pd.Series(data).value_counts()
D 5
A 3
C 2
B 1
dtype: int64

share|improve this answer

answered Mar 7 at 0:56

Marcus LimMarcus Lim

1,405110

add a comment | 

0

Use a defaultdict:

from collections import defaultdict

data = ['A', 'A', 'B', 'A', 'C', 'C', 'A']

d = defaultdict(int)

for element in data:
 d[element] += 1

d # output: defaultdict(int, 'A': 4, 'B': 1, 'C': 2)

share|improve this answer

answered Mar 7 at 1:43

jfaccionijfaccioni

3916

add a comment | 

0

There's not any grouping , just counting, so you can use

from collections import Counter
counter(['A', 'B', 'A'])

share|improve this answer

answered Mar 7 at 3:14

user2699user2699

1,379718

add a comment | 

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0

Use a Series instead.

>>> import pandas as pd
>>> 
>>> data = ['A', 'A', 'A', 'B', 'C', 'C', 'D', 'D', 'D', 'D', 'D']
>>> 
>>> pd.Series(data).value_counts()
D 5
A 3
C 2
B 1
dtype: int64

share|improve this answer

answered Mar 7 at 0:56

Marcus LimMarcus Lim

1,405110

add a comment | 

0

Use a Series instead.

>>> import pandas as pd
>>> 
>>> data = ['A', 'A', 'A', 'B', 'C', 'C', 'D', 'D', 'D', 'D', 'D']
>>> 
>>> pd.Series(data).value_counts()
D 5
A 3
C 2
B 1
dtype: int64

share|improve this answer

answered Mar 7 at 0:56

Marcus LimMarcus Lim

1,405110

add a comment | 

Use a Series instead.

>>> import pandas as pd
>>> 
>>> data = ['A', 'A', 'A', 'B', 'C', 'C', 'D', 'D', 'D', 'D', 'D']
>>> 
>>> pd.Series(data).value_counts()
D 5
A 3
C 2
B 1
dtype: int64

share|improve this answer

answered Mar 7 at 0:56

Marcus LimMarcus Lim

1,405110

>>> import pandas as pd
>>> 
>>> data = ['A', 'A', 'A', 'B', 'C', 'C', 'D', 'D', 'D', 'D', 'D']
>>> 
>>> pd.Series(data).value_counts()
D 5
A 3
C 2
B 1
dtype: int64

share|improve this answer

answered Mar 7 at 0:56

Marcus LimMarcus Lim

1,405110

answered Mar 7 at 0:56

Marcus LimMarcus Lim

1,405110

answered Mar 7 at 0:56

Marcus LimMarcus Lim

1,405110

0

Use a defaultdict:

from collections import defaultdict

data = ['A', 'A', 'B', 'A', 'C', 'C', 'A']

d = defaultdict(int)

for element in data:
 d[element] += 1

d # output: defaultdict(int, 'A': 4, 'B': 1, 'C': 2)

share|improve this answer

answered Mar 7 at 1:43

jfaccionijfaccioni

3916

add a comment | 

0

Use a defaultdict:

from collections import defaultdict

data = ['A', 'A', 'B', 'A', 'C', 'C', 'A']

d = defaultdict(int)

for element in data:
 d[element] += 1

d # output: defaultdict(int, 'A': 4, 'B': 1, 'C': 2)

share|improve this answer

answered Mar 7 at 1:43

jfaccionijfaccioni

3916

add a comment | 

Use a defaultdict:

from collections import defaultdict

data = ['A', 'A', 'B', 'A', 'C', 'C', 'A']

d = defaultdict(int)

for element in data:
 d[element] += 1

d # output: defaultdict(int, 'A': 4, 'B': 1, 'C': 2)

share|improve this answer

answered Mar 7 at 1:43

jfaccionijfaccioni

3916

from collections import defaultdict

data = ['A', 'A', 'B', 'A', 'C', 'C', 'A']

d = defaultdict(int)

for element in data:
 d[element] += 1

d # output: defaultdict(int, 'A': 4, 'B': 1, 'C': 2)

share|improve this answer

answered Mar 7 at 1:43

jfaccionijfaccioni

3916

answered Mar 7 at 1:43

jfaccionijfaccioni

3916

answered Mar 7 at 1:43

jfaccionijfaccioni

3916

0

There's not any grouping , just counting, so you can use

from collections import Counter
counter(['A', 'B', 'A'])

share|improve this answer

answered Mar 7 at 3:14

user2699user2699

1,379718

add a comment | 

0

There's not any grouping , just counting, so you can use

from collections import Counter
counter(['A', 'B', 'A'])

share|improve this answer

answered Mar 7 at 3:14

user2699user2699

1,379718

add a comment | 

There's not any grouping , just counting, so you can use

from collections import Counter
counter(['A', 'B', 'A'])

share|improve this answer

answered Mar 7 at 3:14

user2699user2699

1,379718

from collections import Counter
counter(['A', 'B', 'A'])

share|improve this answer

answered Mar 7 at 3:14

user2699user2699

1,379718

answered Mar 7 at 3:14

user2699user2699

1,379718

answered Mar 7 at 3:14

user2699user2699

1,379718