How to make subset of Counter?

up vote
2
down vote

favorite

I am experimenting with Python standard library Collections.

I have a Counter of things as

>>> c = Counter('achdnsdsdsdsdklaffaefhaew')
>>> c
Counter('a': 4,
 'c': 1,
 'h': 2,
 'd': 5,
 'n': 1,
 's': 4,
 'k': 1,
 'l': 1,
 'f': 3,
 'e': 2,
 'w': 1)

What I want now is to somehow get subset of this counter as another Counter object. Just like this:

>>> new_c = do_subset(c, [d,s,l,e,w])
>>> new_c
Counter('d': 5,
 's': 4,
 'l': 1,
 'e': 2,
 'w': 1)

Thank you in advance.

asked Nov 11 at 2:08

Jay Patel

463

add a comment |

up vote
2
down vote

favorite

I am experimenting with Python standard library Collections.

I have a Counter of things as

>>> c = Counter('achdnsdsdsdsdklaffaefhaew')
>>> c
Counter('a': 4,
 'c': 1,
 'h': 2,
 'd': 5,
 'n': 1,
 's': 4,
 'k': 1,
 'l': 1,
 'f': 3,
 'e': 2,
 'w': 1)

What I want now is to somehow get subset of this counter as another Counter object. Just like this:

>>> new_c = do_subset(c, [d,s,l,e,w])
>>> new_c
Counter('d': 5,
 's': 4,
 'l': 1,
 'e': 2,
 'w': 1)

Thank you in advance.

asked Nov 11 at 2:08

Jay Patel

463

add a comment |

up vote
2
down vote

favorite

I am experimenting with Python standard library Collections.

I have a Counter of things as

>>> c = Counter('achdnsdsdsdsdklaffaefhaew')
>>> c
Counter('a': 4,
 'c': 1,
 'h': 2,
 'd': 5,
 'n': 1,
 's': 4,
 'k': 1,
 'l': 1,
 'f': 3,
 'e': 2,
 'w': 1)

What I want now is to somehow get subset of this counter as another Counter object. Just like this:

>>> new_c = do_subset(c, [d,s,l,e,w])
>>> new_c
Counter('d': 5,
 's': 4,
 'l': 1,
 'e': 2,
 'w': 1)

Thank you in advance.

asked Nov 11 at 2:08

Jay Patel

463

I am experimenting with Python standard library Collections.

I have a Counter of things as

>>> c = Counter('achdnsdsdsdsdklaffaefhaew')
>>> c
Counter('a': 4,
 'c': 1,
 'h': 2,
 'd': 5,
 'n': 1,
 's': 4,
 'k': 1,
 'l': 1,
 'f': 3,
 'e': 2,
 'w': 1)

What I want now is to somehow get subset of this counter as another Counter object. Just like this:

>>> new_c = do_subset(c, [d,s,l,e,w])
>>> new_c
Counter('d': 5,
 's': 4,
 'l': 1,
 'e': 2,
 'w': 1)

Thank you in advance.

python data-structures python-collections

asked Nov 11 at 2:08

Jay Patel

463

asked Nov 11 at 2:08

Jay Patel

463

asked Nov 11 at 2:08

Jay Patel

463

asked Nov 11 at 2:08

Jay Patel

463

asked Nov 11 at 2:08

Jay Patel

463

add a comment |

2 Answers
2

active

oldest

votes

up vote
4
down vote

accepted

You could simply build a dictionary and pass it to Counter:

from collections import Counter

c = Counter('a': 4,
 'c': 1,
 'h': 2,
 'd': 5,
 'n': 1,
 's': 4,
 'k': 1,
 'l': 1,
 'f': 3,
 'e': 2,
 'w': 1)


def do_subset(counter, lst):
 return Counter(k: counter.get(k, 0) for k in lst)


result = do_subset(c, ['d', 's', 'l', 'e', 'w'])

print(result)

Output

Counter('d': 5, 's': 4, 'e': 2, 'l': 1, 'w': 1)

answered Nov 11 at 2:12

Daniel Mesejo

11k1923

2

If you are lazy, you can also pass 'dslew' because strings are iterable.
– jpp
Nov 11 at 2:14

Thank you @daniel it worked.
– Jay Patel
Nov 11 at 2:17

@jpp you made it shorter... awesome...
– Jay Patel
Nov 11 at 2:18

@JayPatel Glad I could help! If my answer helped to solve your problem, please consider marking it as accepted. That's the customary way of indicating that your question is "resolved" and thanking the person who helped you.
– Daniel Mesejo
Nov 11 at 2:20

add a comment |

up vote
0
down vote

You can access each key in c and assign its value to the same key in a new dict.

import collections
c = collections.Counter('achdnsdsdsdsdklaffaefhaew')

def subsetter(c, sub):
 out = 
 for x in sub:
 out[x] = c[x]
 return collections.Counter(out)

subsetter(c, ["d","s","l","e","w"])

Yields:

'd': 5, 'e': 2, 'l': 1, 's': 4, 'w': 1

edited Nov 11 at 2:20

answered Nov 11 at 2:12

Charles Landau

1,7881215

OP wants a Counter not a dict.
– Red Cricket
Nov 11 at 2:19

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
4
down vote

accepted

You could simply build a dictionary and pass it to Counter:

from collections import Counter

c = Counter('a': 4,
 'c': 1,
 'h': 2,
 'd': 5,
 'n': 1,
 's': 4,
 'k': 1,
 'l': 1,
 'f': 3,
 'e': 2,
 'w': 1)


def do_subset(counter, lst):
 return Counter(k: counter.get(k, 0) for k in lst)


result = do_subset(c, ['d', 's', 'l', 'e', 'w'])

print(result)

Output

Counter('d': 5, 's': 4, 'e': 2, 'l': 1, 'w': 1)

answered Nov 11 at 2:12

Daniel Mesejo

11k1923

2

If you are lazy, you can also pass 'dslew' because strings are iterable.
– jpp
Nov 11 at 2:14

Thank you @daniel it worked.
– Jay Patel
Nov 11 at 2:17

@jpp you made it shorter... awesome...
– Jay Patel
Nov 11 at 2:18

@JayPatel Glad I could help! If my answer helped to solve your problem, please consider marking it as accepted. That's the customary way of indicating that your question is "resolved" and thanking the person who helped you.
– Daniel Mesejo
Nov 11 at 2:20

add a comment |

up vote
4
down vote

accepted

You could simply build a dictionary and pass it to Counter:

from collections import Counter

c = Counter('a': 4,
 'c': 1,
 'h': 2,
 'd': 5,
 'n': 1,
 's': 4,
 'k': 1,
 'l': 1,
 'f': 3,
 'e': 2,
 'w': 1)


def do_subset(counter, lst):
 return Counter(k: counter.get(k, 0) for k in lst)


result = do_subset(c, ['d', 's', 'l', 'e', 'w'])

print(result)

Output

Counter('d': 5, 's': 4, 'e': 2, 'l': 1, 'w': 1)

answered Nov 11 at 2:12

Daniel Mesejo

11k1923

2

If you are lazy, you can also pass 'dslew' because strings are iterable.
– jpp
Nov 11 at 2:14

Thank you @daniel it worked.
– Jay Patel
Nov 11 at 2:17

@jpp you made it shorter... awesome...
– Jay Patel
Nov 11 at 2:18

@JayPatel Glad I could help! If my answer helped to solve your problem, please consider marking it as accepted. That's the customary way of indicating that your question is "resolved" and thanking the person who helped you.
– Daniel Mesejo
Nov 11 at 2:20

add a comment |

up vote
4
down vote

accepted

You could simply build a dictionary and pass it to Counter:

from collections import Counter

c = Counter('a': 4,
 'c': 1,
 'h': 2,
 'd': 5,
 'n': 1,
 's': 4,
 'k': 1,
 'l': 1,
 'f': 3,
 'e': 2,
 'w': 1)


def do_subset(counter, lst):
 return Counter(k: counter.get(k, 0) for k in lst)


result = do_subset(c, ['d', 's', 'l', 'e', 'w'])

print(result)

Output

Counter('d': 5, 's': 4, 'e': 2, 'l': 1, 'w': 1)

answered Nov 11 at 2:12

Daniel Mesejo

11k1923

You could simply build a dictionary and pass it to Counter:

from collections import Counter

c = Counter('a': 4,
 'c': 1,
 'h': 2,
 'd': 5,
 'n': 1,
 's': 4,
 'k': 1,
 'l': 1,
 'f': 3,
 'e': 2,
 'w': 1)


def do_subset(counter, lst):
 return Counter(k: counter.get(k, 0) for k in lst)


result = do_subset(c, ['d', 's', 'l', 'e', 'w'])

print(result)

Output

Counter('d': 5, 's': 4, 'e': 2, 'l': 1, 'w': 1)

answered Nov 11 at 2:12

Daniel Mesejo

11k1923

answered Nov 11 at 2:12

Daniel Mesejo

11k1923

answered Nov 11 at 2:12

Daniel Mesejo

11k1923

answered Nov 11 at 2:12

Daniel Mesejo

11k1923

2

If you are lazy, you can also pass 'dslew' because strings are iterable.
– jpp
Nov 11 at 2:14

Thank you @daniel it worked.
– Jay Patel
Nov 11 at 2:17

@jpp you made it shorter... awesome...
– Jay Patel
Nov 11 at 2:18

@JayPatel Glad I could help! If my answer helped to solve your problem, please consider marking it as accepted. That's the customary way of indicating that your question is "resolved" and thanking the person who helped you.
– Daniel Mesejo
Nov 11 at 2:20

add a comment |

2

If you are lazy, you can also pass 'dslew' because strings are iterable.
– jpp
Nov 11 at 2:14

Thank you @daniel it worked.
– Jay Patel
Nov 11 at 2:17

@jpp you made it shorter... awesome...
– Jay Patel
Nov 11 at 2:18

@JayPatel Glad I could help! If my answer helped to solve your problem, please consider marking it as accepted. That's the customary way of indicating that your question is "resolved" and thanking the person who helped you.
– Daniel Mesejo
Nov 11 at 2:20

If you are lazy, you can also pass 'dslew' because strings are iterable.
– jpp
Nov 11 at 2:14

Thank you @daniel it worked.
– Jay Patel
Nov 11 at 2:17

@jpp you made it shorter... awesome...
– Jay Patel
Nov 11 at 2:18

@JayPatel Glad I could help! If my answer helped to solve your problem, please consider marking it as accepted. That's the customary way of indicating that your question is "resolved" and thanking the person who helped you.
– Daniel Mesejo
Nov 11 at 2:20

add a comment |

up vote
0
down vote

You can access each key in c and assign its value to the same key in a new dict.

import collections
c = collections.Counter('achdnsdsdsdsdklaffaefhaew')

def subsetter(c, sub):
 out = 
 for x in sub:
 out[x] = c[x]
 return collections.Counter(out)

subsetter(c, ["d","s","l","e","w"])

Yields:

'd': 5, 'e': 2, 'l': 1, 's': 4, 'w': 1

edited Nov 11 at 2:20

answered Nov 11 at 2:12

Charles Landau

1,7881215

OP wants a Counter not a dict.
– Red Cricket
Nov 11 at 2:19

add a comment |

up vote
0
down vote

You can access each key in c and assign its value to the same key in a new dict.

import collections
c = collections.Counter('achdnsdsdsdsdklaffaefhaew')

def subsetter(c, sub):
 out = 
 for x in sub:
 out[x] = c[x]
 return collections.Counter(out)

subsetter(c, ["d","s","l","e","w"])

Yields:

'd': 5, 'e': 2, 'l': 1, 's': 4, 'w': 1

edited Nov 11 at 2:20

answered Nov 11 at 2:12

Charles Landau

1,7881215

OP wants a Counter not a dict.
– Red Cricket
Nov 11 at 2:19

add a comment |

up vote
0
down vote

You can access each key in c and assign its value to the same key in a new dict.

import collections
c = collections.Counter('achdnsdsdsdsdklaffaefhaew')

def subsetter(c, sub):
 out = 
 for x in sub:
 out[x] = c[x]
 return collections.Counter(out)

subsetter(c, ["d","s","l","e","w"])

Yields:

'd': 5, 'e': 2, 'l': 1, 's': 4, 'w': 1

edited Nov 11 at 2:20

answered Nov 11 at 2:12

Charles Landau

1,7881215

You can access each key in c and assign its value to the same key in a new dict.

import collections
c = collections.Counter('achdnsdsdsdsdklaffaefhaew')

def subsetter(c, sub):
 out = 
 for x in sub:
 out[x] = c[x]
 return collections.Counter(out)

subsetter(c, ["d","s","l","e","w"])

Yields:

'd': 5, 'e': 2, 'l': 1, 's': 4, 'w': 1

edited Nov 11 at 2:20

answered Nov 11 at 2:12

Charles Landau

1,7881215

edited Nov 11 at 2:20

answered Nov 11 at 2:12

Charles Landau

1,7881215

answered Nov 11 at 2:12

Charles Landau

1,7881215

answered Nov 11 at 2:12

Charles Landau

1,7881215

OP wants a Counter not a dict.
– Red Cricket
Nov 11 at 2:19

add a comment |

OP wants a Counter not a dict.
– Red Cricket
Nov 11 at 2:19

OP wants a Counter not a dict.
– Red Cricket
Nov 11 at 2:19

add a comment |

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