['a','a','b','c','c','c']
to
[2, 2, 1, 3, 3, 3]
and
{'a': 2, 'c': 3, 'b': 1}
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1I don't understand the question.Federer– Federer2010-03-08 14:23:46 +00:00Commented Mar 8, 2010 at 14:23
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3It would be most helpful to add at least one line of description.Juergen– Juergen2010-03-08 14:28:51 +00:00Commented Mar 8, 2010 at 14:28
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this question is being asked every day for the last weekSilentGhost– SilentGhost2010-03-08 14:52:06 +00:00Commented Mar 8, 2010 at 14:52
Add a comment
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8 Answers
>>> x=['a','a','b','c','c','c']
>>> map(x.count,x)
[2, 2, 1, 3, 3, 3]
>>> dict(zip(x,map(x.count,x)))
{'a': 2, 'c': 3, 'b': 1}
>>>
4 Comments
Juergen
The result looks pretty, but it has potential O(n^2) runtime behaviour.
chills42
But he never said anything about performance... this solves the problem, if it's too inefficient that's a different problem to solve.
Juergen
That is right -- but did you ever realize what can happen, wenn you run into the O(n^2)-trap? I experienced it, when a simple algorithm killed program performance totally because it happened to be a bigger dataset and the algorithm had such behaviour.
Wim
Relax. You're most likely writing O(n^3) and O(2^n) algorithms all the time, and never even notice them being executed. Solve those problems if they happen.
This coding should give the result:
from collections import defaultdict
myDict = defaultdict(int)
for x in mylist:
myDict[x] += 1
Of course if you want the list inbetween result, just get the values from the dict (mydict.values()).
Comments
On Python ≥2.7 or ≥3.1, we have a built-in data structure collections.Counter to tally a list
>>> l = ['a','a','b','c','c','c']
>>> Counter(l)
Counter({'c': 3, 'a': 2, 'b': 1})
It is easy to build [2, 2, 1, 3, 3, 3] afterwards.
>>> c = _
>>> [c[i] for i in l] # or map(c.__getitem__, l)
[2, 2, 1, 3, 3, 3]
Comments
Use a set to only count each item once, use the list method count to count them, store them in a dict with the item as key and the occurrence is value.
l=["a","a","b","c","c","c"]
d={}
for i in set(l):
d[i] = l.count(i)
print d
Output:
{'a': 2, 'c': 3, 'b': 1}
2 Comments
zach
for me this is the most simple to understand/implement. Do you know if its slower than using the collections Counter?
Mizipzor
For such trivial examples, implement them both and time them. Don't trust my guesses. Or rather, don't care about the speed at all; you're using Python for simplicity, not performance.
Second one could be just
dict(zip(['a','a','b','c','c','c'], [2, 2, 1, 3, 3, 3]))
Comments
For the first one:
l = ['a','a','b','c','c','c']
map(l.count,l)
Comments
d=defaultdict(int)
for i in list_to_be_counted: d[i]+=1
l = [d[i] for i in list_to_be_counted]
