CODEWITHSUNDEEP
python
Chris
Chris

Reputation: 157

Count duplicates in list and assign the sum into list

I have a list with duplicate strings:

lst = ["abc", "abc", "omg", "what", "abc", "omg"]

and I would like to produce:

lst = ["3 abc", "2 omg", "what"]

so basically count duplicates, remove duplicates and add the sum to the beginning of the string.

This is how I do it right now:

from collections import Counter
list2=[]
for i in lst:
  y = dict(Counter(i))
  have = list(accumulate(y.items())) # creating [("omg", 3), ...]

  for tpl in have: #
    join_list = []
    if tpl[1] > 1:
      join_list.append(str(tpl[1])+" "+tpl[0])
    else:
      join_list.append(tpl[0])
  list2.append(', '.join(join_list))

Is there a easier way to obtain the desired result in python?

Upvotes: 4

Views: 642

Answers (5)

Boštjan Mejak
Boštjan Mejak

Reputation: 977

This is the only Pythonic way of doing it, and it is also fast.

import collections

lst = ["abc", "abc", "omg", "what", "abc", "omg"]
duplicates = collections.Counter(lst)

lst = [f"{value} {key}"
       if value > 1 else key
       for (key, value) in duplicates.items()]

Note: this code only works with Python 3.6+ because of the f-string syntax in the list comprehension.

Upvotes: 0

Marcel
Marcel

Reputation: 3258

Another possible solution with comments to help...

import operator

#list
lst = ["abc", "abc", "omg", "what", "abc", "omg"]

#dictionary
countDic = {}

#iterate lst to populate dictionary: {'what': 1, 'abc': 3, 'omg': 2}
for i in lst:
    if i in countDic:
        countDic[i] += 1
    else:
        countDic[i] = 1

#clean list
lst = []

#convert dictionary to an inverse list sorted by value: [('abc', 3), ('omg', 2), ('what', 1)]
sortedLst = sorted(countDic.items(), key=operator.itemgetter(0))

#iterate sorted list to populate list
for k in sortedLst:
    if k[1] != 1:
        lst.append(str(k[1]) + " " + k[0])
    else:
        lst.append(k[0])

#result
print lst

Output:

['3 abc', '2 omg', 'what']

Upvotes: 1

Vermillion
Vermillion

Reputation: 1308

Try this:

lst = ["abc", "abc", "omg", "what", "abc", "omg"]
l = [lst.count(i) for i in lst] # Count number of duplicates
d = dict(zip(lst, l)) # Convert to dictionary
lst = [str(d[i])+' '+i if d[i]>1 else i for i  in d] # Convert to list of strings

Upvotes: 1

Prune
Prune

Reputation: 77837

You've properly used the Counter type to accumulate the needed values. Now, it's just a matter of a more Pythonic way to generate the results. Most of all, pull the initialization out of the loop, or you'll lose all but the last entry.

list2 = []
for tpl in have:
    count = "" if tpl[1] == 0 else str(tpl[1])+" "
    list2.append(count + tpl[0])

Now, to throw all of that into a list comprehension:

list2 = [ ("" if tpl[1] == 0 else str(tpl[1])+" ") + tpl[0] \
          for tpl in have]

Upvotes: 1

juanpa.arrivillaga
juanpa.arrivillaga

Reputation: 95948

It seems you are needlessly complicating things. Here is a very Pythonic approach:

>>> import collections
>>> class OrderedCounter(collections.Counter, collections.OrderedDict):
...   pass
... 
>>> lst = ["abc", "abc", "omg", "what", "abc", "omg"]
>>> counts = OrderedCounter(lst)
>>> counts
OrderedCounter({'abc': 3, 'omg': 2, 'what': 1})
>>> ["{} {}".format(v,k) if v > 1 else k for k,v in counts.items()]
['3 abc', '2 omg', 'what']
>>>

Upvotes: 7

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