'Code in Python that count repeated elements and return the item repeated most times and how many times

I have a list with integers, I need to find which item repeat the most, then the function should return two variables (value, number) value is the repeated element and number is the times that repeated

input

B = [1, 5, 1, 3, 5, 1]

output

1 3

I got a code that works with the first item in the list, but when it goes to the next loop it start everything in zero, so I was struggling to store what I got in the first loop and compare with the second loop to see which one I need to keep.

For this function I am not allowed to use any module or library like "count" or any other advanced code, just as simple as it can be for a rookie level

def counting(B):
   n = len(B)
   tempcount = []
   tempcval = []

   for i in range(0,n-1):
      cval=B[i]
      count = 0

      for next in B:
         if next == cval:
              count += 1

         tempcount.append(count)
         tempcval.append(cval)

      return count,cval

def main():
   list = [1, 5, 1, 3, 5, 1]
   a,b = counting(list)
   print(a,b)

main()

expected output

3 1

actual output

2 5
python-3.xlistfor-looppycharmcounting


Solution 1:[1]

A good solution could be to use collections.Counter:

>>> import collections
>>> c = collections.Counter([1, 5, 1, 3, 5, 1])
>>> c
Counter({1: 3, 5: 2, 3: 1})
>>> c.most_common(1)
[(1, 3)]

In a function this could look like this:

import collections

def counting(the_list):
    c = collections.Counter(the_list)
    data = c.most_common(1)[0]
    print('value', data[0])
    print('count', data[1])

    return data[1], data[0]

Solution 2:[2]

Since I'm familiar with pandas, dataframe manipulating library

If I were you, I will use

import pandas as pd
temp = pd.Series([1, 5, 1, 3, 5, 1])
temp2 = temp.value_counts()
temp2.index[0], temp2.iloc[0]

But the first answer seems more simple hahaha

Solution 3:[3]

A solution using nothing requiring any imports. Your code had some issues in that you never actually used those tempcount and tempcvals. You'd append to them, but nothing else. I took them along for the ride this time and am using them to store the counts of each value, basically turning tempcval into a non duplicate list of the inputted list.

def counting(B):
   n = len(B)
   tempcount = []
   tempcval = []

   for next in B:
      if next in tempcval:
         point = tempcval.index(next)
         tempcount[point] = tempcount[point] + 1
      else:
         tempcval.append(next)
         tempcount.append(1)

   maxCount = max(tempcount)
   point = tempcount.index(maxCount)

   return tempcount[point],tempcval[point]

def main():
   list = [1, 5, 1, 3, 5, 1]
   a,b = counting(list)
   print(a,b)

main()

So there is one issue with this code. It works in the example given, but say you gave 

list = [1, 5, 1, 3, 5, 1, 5]

the code would still return 3 1 since the 1 appeared before the 5.

Solution 4:[4]

This question looks to be solved, but here's a spin using the collections.Counter method:

from collections import Counter

my_list = [1, 5, 1, 3, 5, 1, 5]
my_list_counted = Counter(my_list)
most_common_items = my_list_counted.most_common(len(my_list_counted))
print(most_common_items[0][0], most_common_items[-1][0])

>>> 1 3

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 Ralf
Solution 2 Noppu
Solution 3
Solution 4 irahorecka

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