+ 1

# Loop question, I've tried everything that I knew I just don't know. Please help me solve it out

How to I make a code that gives you how many are same in a list, using a loop, for example: Num = [1, 3, 2, 5, 7, 8, 7, 3, 1, 1] And it returns: 1 = 3 2 = 1 3 = 3 5 = 1 7= 2 8 = 1

2nd Feb 2023, 5:56 AM
sunshine + 7
svnshqinz , here is a very basic approach without dictionary > since numbers can occur multiple times in the input values, we have to take care that only the first occurrence of each value is counted. to do so, we need a helpers list *already_seen*, where we can append numbers that have been already counted. > we start a for loop, that gets one number at a time from the input list and is stored in a loop variable variable like *num* > inside the loop, we check if the current number is NOT already seen. this can be done with an *if* conditional and using the *in* operator. > if the current number is not in *already_seen*, we can output the result with print(...) to get the count of the specific number, we can use the string.method count(...) like: ... nums.count(num) > after this, we have to append the current number (variable *num*) to list *already_seen* > if the loop is terminated, all numbers with the counts have been printed
2nd Feb 2023, 12:14 PM
Lothar + 6
another thing i wanted to mention: it is not seen as very helpful when we are going to post a ready-made code, as long as the op has not shown his attempt here. it is more helpful to give hints and tips, so that the op has a chance to find a solution by himself.
2nd Feb 2023, 12:20 PM
Lothar + 3
Should the output be in sorted form like that 1 = 3 ... 8 = 1?
2nd Feb 2023, 6:18 AM
Ipang
+ 3
svnshqinz Create a set from <Num> list to get only unique elements name it <num_set> Then create a list named <sorted_num> from <num_set> and sort the list Next, create a dictionary named <num_freq> where each item key is the value obtained from <sorted_num>, and item value is the frequency of the value in <Num> list. Lastly, display the numbers and their frequency you have collected in <num_freq> dictionary You can try this for now, or ask others how their code example work : )
2nd Feb 2023, 6:46 AM
Ipang
+ 2
Ipang yes
2nd Feb 2023, 6:36 AM
sunshine + 2
Ipang thank you! Actually it didnt need to be sorted but I was to distracted and didn't notice :)
2nd Feb 2023, 8:04 AM
sunshine + 1
print(list(set([num.count(a) for a in num])))
2nd Feb 2023, 6:09 AM
Jay Matthews + 1
Num = [1, 3, 2, 5, 7, 8, 7, 3, 1, 1] print(*[f'{n} = {Num.count(n)}' for n in sorted(set(Num))],sep='\n')
2nd Feb 2023, 6:27 AM
Steve + 1
Lothar, ok thank you very much :)
2nd Feb 2023, 4:59 PM
sunshine 