How do you check if a list's item is equal to another item in the same list

Question

I understand the basics of this problem however I need help on how I can do this the most efficient way possible (taking the least amount of time for the programmer however not substituting stability of the code or efficiency).

Let's say we have a string:

grades=str(input("Enter a string"))

in my code, I would join a space between all characters in the string above and then split the characters into separate items in the same list:

grades=" ".join(grades)
grades.split(" ")

I then want to use loops of some sort to search the list for repeating items. However, I want to learn how I can do this the most efficient way possible:

x=len(grades)
for i in range(0, x):
    if grades[i] ==  # here is were I'm having trouble

I want to know how I can search whether 1 item in the list is equal to any item in the whole list itself. Kind regards.

Mehrdad Pedramfar · Accepted Answer · 2018-11-10 11:01:55Z

4

I make an example:

from collections import Counter

a =[1,2,3,4,1,2]
c = Counter(a)
for k,v in c.items():
    if v>1:
        print(k,'repeated more than once')

Here the c will be a Counter object like this Counter({1: 2, 2: 2, 3: 1, 4: 1}). the keys are the array values and values are the count of them. So I write the for for your understanding. You can do anything with c, it acts like a dict.

 >> [k for k,v in c.items() if v>1]
 [1, 2]

edited Nov 10, 2018 at 11:01

answered Nov 10, 2018 at 10:51

Mehrdad Pedramfar

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3 Comments

Coder Cody Over a year ago

Thank you however once tested I found an error, it is much easier if you do not use both k and v in the 4th line of code. You can just use k or v, not both.

quamrana Over a year ago

I get TypeError: 'int' object is not iterable for the line: for k,v in c:.

Mehrdad Pedramfar Over a year ago

sorry I forgot .items(). it is fixed now.@quamrana

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