How to write an if condition such that python should execute if statement if the element in a listcan be converted to type int [duplicate]

Question

How can i write the code for an if condition where i want to see if i can convert the type of an element in a list to int. If its possible, then if condition should be executed.Otherwise, it should go to else statement

For eg:

lst=['@','#','^','%','20']

''' Now i am going to iterate through the list and check if i can convert the elements  to type int. If it can be converted then i know there is a number in the list and i do some operations on the number'''

for ele in lst:
    if(type(int(ele)=='int'):
        #do some operation on the number eg:20/2 and print the value
    else:
        continue

Please give me an idea as to how to write the if condition.

You should instead try to convert the element to int and proceed if successful, except that you do something else when that fails with a ValueError. — Karl Knechtel
– Karl Knechtel, Commented Nov 15, 2020 at 17:09
Does this answer your question? Python: Test if value can be converted to an int in a list comprehension — Mickey
– Mickey, Commented Nov 15, 2020 at 17:14

Christian K. · Accepted Answer · 2020-11-15 17:19:05Z

1

Try to convert to int and catch the error:

for ele in lst:
    try:
        val = int(ele)
    except ValueError:
        continue
    else:
        print(val)

answered Nov 15, 2020 at 17:19

Christian K.

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Comments

Sash Sinha · Accepted Answer · 2020-11-15 17:33:01Z

0

You could check if each element is an int using a try block in a helper function is_int:

def is_int(s):
    try: 
        int(s)
        return True
    except ValueError:
        return False

lst = ['@', '#', '^', '%', '20']
for s in lst:
    if is_int(s):
        int_s = int(s)
        print(f'{int_s}/2 = {int_s/2}')
    else:
        continue

Output:

20/2 = 10.0

edited Nov 15, 2020 at 17:33

answered Nov 15, 2020 at 17:15

Sash Sinha

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Comments

avarice · Accepted Answer · 2020-11-15 17:09:51Z

0

Try with .isdigit()+a list comprehension (perfect couple ;-):

lst=['@','#','^','%','20']
[str(x).isdigit() for x in lst]

Caveat: does not work for float strings, for floats:

[str(x).replace('.','').isdigit() for x in lst]

answered Nov 15, 2020 at 17:09

avarice

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Comments

IceJonas · Accepted Answer · 2020-11-15 17:26:09Z

0

I think what you are looking for is :

if isinstance(ele,int)

lst = ['@', '#', '^', '%', 20]

for ele in lst:
    if isinstance(ele,int):
        print("This is int: " + str(ele))
    else:
        print("This is not int: " + ele)

Very simple and easy.

answered Nov 15, 2020 at 17:26

IceJonas

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2 Comments

loginator Over a year ago

Thank you. But I am checking if the elements can be converted to int type. not if the element is of type int.

IceJonas Over a year ago

Oh sorry about that :)

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