I'm a python newbie, so please bear with me.
I need to find the frequency of elements in a list
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
output->
b = [4,4,2,1,2]
Also I want to remove the duplicates from a
a = [1,2,3,4,5]
views:
455answers:
7
+6
A:
For your first question, iterate the list and use a dictionary to keep track of an elements existsence.
For your second question, just use the set operator.
David in Dakota
2010-01-29 12:10:59
Can you please elaborate on the first answer
Bruce
2010-01-29 12:14:59
+10
A:
To count the number of appearances:
from collections import defaultdict
appearances = defaultdict(int)
for curr in a:
appearances[curr] += 1
To remove duplicates:
a = set(a)
Idan K
2010-01-29 12:16:03
+1 for collections.defaultdict. Also, in python 3.x, look up collections.Counter. It is the same as collections.defaultdict(int).
hughdbrown
2010-01-29 13:54:30
+10
A:
Counting the frequency of elements is probably best done with a dictionary:
b = {}
for item in a:
b[item] = b.get(item, 0) + 1
To remove the duplicates, use a set:
a = list(set(a))
lindelof
2010-01-29 12:16:19
@phkahler: Mine would only a tiny bit better than this. It's hardly worth my posting a separate answer when this can be improved with a small change. The point of SO is to get to the *best* answers. I could simply edit this, but I prefer to allow the original author a chance to make their own improvements.
S.Lott
2010-01-29 16:58:48
+2
A:
seta = set(a)
b = [a.count(el) for el in seta]
a = list(seta) #Only if you really want it.
Lakshman Prasad
2010-01-29 12:18:15
using lists `count` is ridiculously expensive and uncalled for in this scenario.
Idan K
2010-01-29 12:20:52
+12
A:
Since the list is ordered you can do this:
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
from itertools import groupby
[len(list(group)) for key, group in groupby(a)]
Output:
[4, 4, 2, 1, 2]
Nadia Alramli
2010-01-29 12:18:39
nice, using `groupby`. I wonder about its efficiency vs. the dict approach, though
Eli Bendersky
2010-01-29 12:20:27
@Eli, yeah I'm not sure about its efficiency. But it doesn't hurt to have a variety of solutions.
Nadia Alramli
2010-01-29 12:28:51
The python groupby creates new groups when the value it sees changes. In this case 1,1,1,2,1,1,1] would return [3,1,3]. If you expected [6,1] then just be sure to sort the data before using groupby.
Evan
2010-01-30 22:41:57
I wonder if there's a way to skip the conversion to a list in `len(list(group))`.
Cristian Ciupitu
2010-03-22 12:17:46
+2
A:
In Python 2.7+, you could use collections.Counter to count items
>>> a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
>>>
>>> from collections import Counter
>>> c=Counter(a)
>>>
>>> c.values()
[4, 4, 2, 1, 2]
>>>
>>> c.keys()
[1, 2, 3, 4, 5]
S.Mark
2010-01-29 13:00:53
+5
A:
In Python 2.7, you can use collections.Counter:
import collections
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
counter=collections.Counter(a)
print(counter)
# Counter({1: 4, 2: 4, 3: 2, 5: 2, 4: 1})
print(counter.values())
# [4, 4, 2, 1, 2]
print(counter.keys())
# [1, 2, 3, 4, 5]
print(counter.most_common(3))
# [(1, 4), (2, 4), (3, 2)]
If, like me, you are using Python 2.6 or older, you can download it here.
unutbu
2010-01-29 13:02:41