Check if there is at least one element starting with specific character in a list

Question

I have a list containing a group of numbers ( str ) , and I want to check if there is at least one element in the list starting with '+' .

The problem with my code is that an infinite loop will occur if no number starts with '+' , I need a better solution?

my_list = ['11112352', '222222003', '5682052003', '21543003', '98756003', '+004454883']
while True:
    for number in my_list:
        if number.startswith('+'):
            break

Answer 1

my_list = ['11112352', '222222003', '5682052003', '21543003', '98756003', '+004454883']
for number in my_list:
    if number.startswith('+'):
        break

This:

while True:

is unnecessary, the for loop will already check every item in the list

To make it clearer, if it print every iteration:

my_list = ['11112352', '222222003', '5682052003', '21543003', '98756003', '+004454883']
for number in my_list:
    print(number)
    if number.startswith('+'):
        break

I get

11112352
222222003
5682052003
21543003
98756003
+004454883

so as you can see the while loop is unnecessary

Answer 2

Here's a simpler solution:

my_list = ['11112352', '222222003', '5682052003', '21543003', '98756003', '+004454883']
result = [x for x in my_list if x.startswith('+')]
if result:
    # do something if result is not empty
else:
    # result is empty -- no '+' entries in my_list

Answer 3

If you just need a True / False result, you can try this:

any(number for number in my_list if number[0] == "+")

Function any delivers True if at least one element of the list passed as argument is True .

Edited as suggested by Ann Zen . See also How exactly any works

Answer 4

To check if any element starts with a '+' , use builtin functions any and str.startswith . You can nearly write down the exact words as Python code:

starts_with_plus = any(element.startswith('+') for element in my_list)

This will return a boolean value. This will also return on the first instance where the condition is True , like returning out of a loop, making this very efficient.

If you want to check for other prefixes as well, just put it in a small function:

def starts_with(lst, prefix):
    return any(element.startswith(prefix) for element in lst)

starts_with_plus = starts_with(my_list, '+')

Answer 5

Use the any method:

my_list = ['11112352', '222222003', '5682052003', '21543003', '98756003', '+004454883']
print(any(s[0] == '+' for s in my_list))

Output:

True

Answer 6

You can do this with the help of a for loop and a flag variable:

flag = 0

for item in my_list:
    if item.startswith('+'):
        flag = 1
        break

print(flag)

If you get an output of 0 , then it's obvious that no element starts with a + . On the other hand, if the output of the print statement is 1 , there's at least one element that starts with a + .