['a','a','b','c','c','c']
to
[2, 2, 1, 3, 3, 3]
and
{'a': 2, 'c': 3, 'b': 1}
+3
A:
a = ['a','a','b','c','c','c']
b = [a.count(x) for x in a]
c = dict(zip(a, b))
I've included Wim answer. Great idea
luc
2010-03-08 14:23:38
`c = dict(zip(a, b))` should do it for computing `c`
Wim
2010-03-08 14:25:19
+3
A:
Second one could be just
dict(zip(['a','a','b','c','c','c'], [2, 2, 1, 3, 3, 3]))
Wim
2010-03-08 14:24:19
+19
A:
>>> 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}
>>>
S.Mark
2010-03-08 14:24:55
But he never said anything about performance... this solves the problem, if it's too inefficient that's a different problem to solve.
chills42
2010-03-08 14:48:24
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.
Juergen
2010-03-08 15:29:00
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.
Wim
2010-03-10 20:54:38
+5
A:
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()).
Juergen
2010-03-08 14:25:09
+1
A:
d=defaultdict(int)
for i in list_to_be_counted: d[i]+=1
l = [d[i] for i in list_to_be_counted]
starhusker
2010-03-08 14:25:56
+4
A:
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}
mizipzor
2010-03-08 14:30:19
+1
A:
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]
KennyTM
2010-08-16 19:27:05