python problem : print the frequencies of each number

Question

I have to solve this problem like below manner only, As you see in output, The only wrong thing in output I get is, 4 and 5 are count two times, How I can get rid of this, please improvise below code only. do not change the code completely.

n=[1,2,3,4,4,5,5]
i=0
a=0
count=0
while i<len(n):
  a=0
  count=0
  while a<len(n):
    if n[i]==n[a]:
      count=count+1
    a=a+1
  print(n[i],"present",count,"times")
  i=i+1

output:

1 present 1 times
2 present 1 times
3 present 1 times
4 present 2 times
4 present 2 times
5 present 2 times
5 present 2 times

I see you want to do it that way only. Then include a condition before print: `if i > 0 and n[i] != n[i-1]`: — SomeDude, Aug 16 '22 at 15:25

score 1 · Answer 1 · answered Aug 16 '22 at 15:25

you can use a Counter to do this efficiently https://docs.python.org/3/library/collections.html#collections.Counter

from collections import Counter

n = [1,2,3,4,4,5,5]

c = Counter(n)

for val, count in c.items():
   print(f"{val} present {count} times")

prints:

1 present 1 times
2 present 1 times
3 present 1 times
4 present 2 times
5 present 2 times

score 1 · Answer 2 · answered Aug 16 '22 at 15:27

1

I purpose to use set to do so. Does it suit your needs?

n_list = [1,2,3,4,4,5,5]
n_set = set(n_list)
for i in n_set:
  print(i, " present ", n_list.count(i), "times")

answered Aug 16 '22 at 15:27

Markus Wanx

54
5

why the minus. This looks good to me – resigned Aug 16 '22 at 15:33
@resigned The question asks for help with their code and says to *"not change the code completely"*. – Kelly Bundy Aug 16 '22 at 15:34
My bad, sorry. I tried to figure out some easy solution and missed above mentioned information. – Markus Wanx Aug 16 '22 at 15:46
also the set will not preserve the ordering – Anentropic Aug 16 '22 at 16:16

score 1 · Answer 3 · answered Aug 16 '22 at 15:28

Sticking to your original code and not the other many ways to solve this using libraries and dictionaries and whatnot.

You can check to see if you the item you are counting has already occurred in the list and skip processing it.

n=[1,2,3,4,4,5,5]
i=0
a=0
count=0
while i<len(n):
  a=0
  count=0
  
  #check if this value has already been encountered in the list
  if n[i] not in n[:i]: 
    while a<len(n):
        if n[i]==n[a]:
            count=count+1
        a=a+1
    print(n[i],"present",count,"times")
  i=i+1

If that's too advanced (in the event this is homework). You could create a second list to keep track of the values you've already checked:

n=[1,2,3,4,4,5,5]
i=0
a=0
count=0
n2=[]
while i<len(n):
  a=0
  count=0
  if n[i] not in n2: 
    while a<len(n):
        if n[i]==n[a]:
            count=count+1
        a=a+1
    print(n[i],"present",count,"times")
    n2.append(n[i])
  i=i+1

score 0 · Answer 4 · answered Aug 16 '22 at 17:14

in general we should use a for loop if we iterate over a sequence because it's length is known. doing so avoids the need of helper variables. the code for this approach looks more clean and readable (2)

to get rid of the problem of duplicating outputs, we can use a list like seen (1). if a number is not in seen (3), we can print it out. to get the total number of occurrences of the number, we use nums.count(nums) (4). in the step after this, we add the current number used to the list 'seen' (5). if the number is already in the list 'seen', the loop will take the next number from the input list.

nums = [1,2,3,4,4,5,5]
seen = [] # (1)
for num in nums: # (2)
    if num not in seen: # (3)
        print(num,"present",nums.count(num),"times") # (4)
        seen.append(num) # (5)

python problem : print the frequencies of each number

4 Answers4