CODEWITHSUNDEEP
pythoncountduplicates
nimbusmom
nimbusmom

Reputation: 11

How to remove duplicate prints in Python?

So I wanted to write a program that would print out how many times a number is repeated in a list. I'm very new to Python and I used the .count() command to get the number of times an element is repeated. I want the output to follow the format "The number n is repeated n times".

This is what I have so far:

x = [1,1,2,4,5,7,2,3,3,8,2,3,6,7]
index = 0 
while index < len(x):
print('The number {} is repeated {} times.'.format(x[index],x.count(x[index])))
index = index + 1

And this is the output:

The number 1 is repeated 2 times
The number 1 is repeated 2 times
The number 2 is repeated 3 times
The number 4 is repeated 1 times
The number 5 is repeated 1 times
The number 7 is repeated 2 times
The number 2 is repeated 3 times
The number 3 is repeated 3 times
The number 3 is repeated 3 times
The number 8 is repeated 1 times
The number 2 is repeated 3 times
The number 3 is repeated 3 times
The number 6 is repeated 1 times
The number 7 is repeated 2 times

I want the output to show how many times a number is repeated only once, and in ascending order. How can I make the output come out like this:

The number 1 is repeated 2 times.
The number 2 is repeated 3 times. 
The number 3 is repeated 3 times. 
...

Thank you! :)

Upvotes: 0

Views: 628

Answers (5)

Jan Christoph Terasa
Jan Christoph Terasa

Reputation: 5935

If you want to emulate what Counter does, you can just fill your own dict. I intentionally do not use try and except, or setdefault, or defaultdict, to show that the basic logic is very simple, just a loop and an if-else block:

def count_elements(lst):
    ret = {}
    for elem in lst:
        if elem in ret:
            ret[elem] += 1
        else:
            ret[elem] = 1
    return ret

def print_counts(counts):
    for n, c in sorted(counts.items()):
        print(f'The number {n} is repeated {c} times')

x = [1,1,2,4,5,7,2,3,3,8,2,3,6,7]
c = count_elements(x)
print_counts(c)

Upvotes: 0

solid.py
solid.py

Reputation: 2812

This counts all occurrences of numbers and prints them in ascending order.

from collections import Counter

x = [1, 1, 2, 4, 5, 7, 2, 3, 3, 8, 2, 3, 6, 7]

for key, val in sorted(Counter(x).items()):
    print(f'The number {key} is repeated {val} times.')

Code Output:

The number 1 is repeated 2 times.
The number 2 is repeated 3 times.
The number 3 is repeated 3 times.
The number 4 is repeated 1 times.
The number 5 is repeated 1 times.
The number 6 is repeated 1 times.
The number 7 is repeated 2 times.
The number 8 is repeated 1 times.

Upvotes: 1

JolonB
JolonB

Reputation: 428

You could try this (though any of these answers seem just as valid):

x = [1,1,2,4,5,7,2,3,3,8,2,3,6,7]
x_sorted = sorted(set(x)) # remove duplicates with set, convert to list, and sort
for num in x_sorted:
    print('The number {} is repeated {} times.'.format(num,x.count(num)))

This may not be the prettiest solution (I'm not a fan of the sorted(set(x)) line), but it prints the results in ascending order.

This outputs:

The number 1 is repeated 2 times.
The number 2 is repeated 3 times.
The number 3 is repeated 3 times.
The number 4 is repeated 1 times.
The number 5 is repeated 1 times.
The number 6 is repeated 1 times.
The number 7 is repeated 2 times.
The number 8 is repeated 1 times.

Upvotes: 0

deceze
deceze

Reputation: 522451

Use a Counter to do the counting and "deduplicating", then iterate over its results:

from collections import Counter

x = [1,1,2,4,5,7,2,3,3,8,2,3,6,7]

for n, c in Counter(x).items():
    print('The number {} is repeated {} times.'.format(n, c))

For printing ordered by least common to most common:

for n, c in reversed(Counter(x).most_common()):
    ...

For printing ordered by the values:

for n, c in sorted(Counter(x).items()):
    ...

Upvotes: 2

Shadowcoder
Shadowcoder

Reputation: 972

Try this:

x = [1,1,2,4,5,7,2,3,3,8,2,3,6,7]
index = 0 
while index < len(x):
    if index==x.index(x[index]):
        print('The number {} is repeated {} times.'.format(x[index],x.count(x[index])))
    index = index + 1

Upvotes: 0

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