# Count how many times values has appeared in list so far

I have a list of non-unique values. I want to output a list of the same length where each value corresponds to how many times that value has appeared so far.

The code I have written to do this seems over-complicated, so I think there has to be a better way.

x = [1,1,2,3,1,3]

def times_so_far(ls):

out = [0]*len(x)
counted = {}
for i,v in enumerate(x):
if v in counted:
counted[v] += 1
else:
counted[v] = 0

out[i] = counted[v]
return out

times_so_far(x)
#[0, 1, 0, 0, 2, 1]

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I have rolled back the last edit. Please see What to do when someone answers. – Ethan Bierlein Jan 29 at 18:23
What you wrote is actually pretty good, certainly better than the answer you accepted. – user2357112 Jan 29 at 22:07

Maybe using list.count() you can achieve the same with less lines?

x = [1,1,2,3,1,3]

def times_so_far(ls):

out = [0]*len(ls)
for i in xrange(len(ls)):
out[i] = ls[:i].count(ls[i])
return out


This can be written with list comprehension as mentioned by Caridorc, and it removes the first len(ls) call this way (though maybe at the cost of resizing out for every element of the list):

def times_so_far(ls):

return [ls[:i].count(ls[i]) for i in xrange(len(ls))]


Now, if you're going to work with lists of positive integers of a small value, I would recommend the following, as it's similar to yours but it works with indices instead of dict keys:

def times_so_far(ls, maxValue):
temp = [0]*(maxValue+1)
out = [0]*len(ls)
for i in xrange(len(ls)):
out[i] = temp[ls[i]]
temp[ls[i]] += 1
return out

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Yes, list.count seems much better. I think I'll use this (I think it's ls[i] and not ls[i+1], though) – C_Z_ Jan 29 at 18:20
Isn't this slower, and also have the same worst case memory usage? – Joe Wallis Jan 29 at 18:30
@JoeWallis It's Python... probably the slowest language out there. If performance is a problem then I would suggest moving on to another language. – hackerman Jan 29 at 18:38
My point is your algorithm's speed is $O(n^2)$ where OP's was $O(n)$. And you also use $O(3n)$ memory. IMO that's a bad thing to say is a good algorithm. In any language. – Joe Wallis Jan 29 at 18:45
@hackerman out and ls is 2n. ls[:i] is a copy which is why i[:] is good. And so $O(3n)$. And dict lookup is $O(1)$. Since this is tagged algorithm I would assume they want it improved... – Joe Wallis Jan 29 at 18:53

You can use a defaultdict, to remove the if v else. To achieve this, you can do:

from collections import defaultdict
counted = defaultdict(int)
for i,v in enumerate(x):
counted[v] += 1


To remove the need for out and enumerate you can yield on each iteration.

from collections import defaultdict

def times_so_far(list_):
counted = defaultdict(int)
for v in list_:
counted[v] += 1
yield counted[v]

print(list(times_so_far([1,1,2,3,1,3])))
#[0, 1, 0, 0, 2, 1]


It will return a generator object, hence why I use list. If this is a problem then you can change yield to out.append() and re-add out.

This has $O(n)$ runtime, with a min $O(n)$ and maximum $O(2n)$ memory usage.

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I would use the "high-performance" (as described in documentation) Counter class for this task. You can actually get a complete counter at every step, but that is of course more than you asked for.

This is my implementation:

from collections import Counter

def times_so_far(nums):
return list(times_so_far_generator(nums))

def times_so_far_generator(nums):
counter = Counter()
for num in nums:
counter[num] += 1
yield counter[num]


I like this solution because:

• It is fast: uses generators + Counter and asymptotically optimal $O(n)$ performance with respect to both memory and time.
• It is very readable and understandable
• It generalizes (yield counter instead of yield counter[num] to get a full Counter at every position). You can also return a tuple of the counts with the element with yield (num, counter[num])
• It uses the specifically designed Counter instead of a generic defaultdict, and so is more fitting to the task (and can possibly perform better, as values are integers instead of any object)
• It works on arbitrary iterables (even infinitely long ones, if you use the generator version)
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Whoever down voted...please explain yourselves? – mleyfman Jan 29 at 20:35
Not the downvoter, but calling Counter "high-performance" may be something of a misnomer. The only high-performance aspects of it relative to an ordinary dict or a defaultdict are creating or update-ing it from an input iterable, and even then, only on Python 3.3 and up. Other than that, it's either equivalent in speed or slower than dict or defaultdict. – user2357112 Jan 29 at 22:22
@user2357112 I was simply referring to the way it is described in the official documentation. The collections module is described as "High-performance container datatypes" – mleyfman Jan 29 at 22:34
I timed both Counter and defaultdict in Python2 and Python3. (It's only one data set so it may be significantly different on a larger data set.) – Joe Wallis Jan 29 at 23:54

I propose a very slow solution, $O(N^2)$ in time and space.

The benefits of this solution are its clarity and simplicity.

Knowing that:

• xs[:i] gives all the elements of xs up to the index i
• enumerate yields an index, value pair,

The following is easier to understand than English:

def times_so_far(xs):
return [xs[:index].count(x) for index, x in enumerate(xs)]


In case you care about performance, you may prefer using a Counter or a defaultdict as suggested by @Joe Wallis and @Ethan Bierlein

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You're right - there is a better way. Python lists have a default built-in function named count. For example, if I wanted to count the number of times the string "lol" occurred in a list, I could just do this:

my_list = [1, 2, "lol", 3, 4, "lol", 5, 6, 7, 8, "lol"]
print(my_list.count("lol"))


This is ultimately better than generating a list of counts corresponding to each value as well, since it's less memory-intensive and much easier to work with.

In addition, if you want to take advantage of the powerful collections module, you should add this to the top of your code:

from collections import Counter


Then you can use the Counter object like this:

>>> print(Counter([1, 2, "lol", 3, 4, "lol", 5, 6, 7, 8, "lol"]))
Counter({'lol': 3, 1: 1, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1, 7: 1, 8: 1})


Hope this helps! :)

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