Python – check if a letter is in a list

Question

If a letter (string) is in a list,
find_letter([‘o’, [‘hello’, ‘c’, ‘bye’]),
return True, if not return False.

def find_letter(lst):

    lst=['o','hello', 1]
    n='o'

    if not lst:          
        return 0

    elif lst[0] == n:
        return True

    elif find_letter(lst[0:]):
        return True

    else: 
        return False


print(find_letter(lst))

It does return ‘True’, but I am not sure if this is the right way to do this. Maybe there’s a better way? In the second elif-statement, is python going through all the elements in the list if the first one doesn’t contain the letter? The function must be recursive.

Asked By: raspberrysupreme

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Source

Answer 1

I think the most pythonic approach would be to use

def find_letter(letter, lst):
    return any(letter in word for word in lst)

The beauty of this is that it iterates over lst and returns as soon as one of the words in that list contains letter. Also, it doesn’t need to recurse.

This returns False instead of 0 if lst is empty, though (unlike your program) but since False evaluates to 0 anyway (and vice versa), that’s not really an issue.

Answered By: Tim Pietzcker

Answer 2

You want to find the Membership

Please refer this code to resolve your problem.
Suppose

list1 = ['physics', 'chemistry', 1997, 2000];
list2 = [1, 2, 3, 4, 5, 6, 7 ];

print "list1[0]: ", list1[0]
print "list2[1:5]: ", list2[1:5]
print 3 in list2

Output:

list1[0]:  physics
list2[1:5]:  [2, 3, 4, 5]
True

Answered By: Sohair Ahmad

Answer 3

Just realized OP wants to check A string only
You can define a function and match the string in a list recursively like this:

def plist(lst, n):
    # loop inside the list
    for each in lst:
        # if "each" is also a list (nested), check inside that list, too
        if isinstance(each, list):
            # this will reuse your "plist" and loop over any nested list again
            plist(each, n)
        # if n matches any element in the list, return True
        if any(n in each_letter for each_letter in each):
            return True
    # if after looping over your list + nested list, return False if no match is find
    else:
        return False

>> lst = ['o', ['hello', 'c', 'bye']]
>> plist(lst, 'o')
>> True
>> plist(lst, 'h')
>> True
>> plist(lst, 'z')
>> False

Hope this solves your problem.

Answered By: Anzel

Answer 4

Here is your error

def find_letter(lst):  # You receive your list as lst
    lst=['o','hello', 1]  # Opppsss! You override it. It does not matter what you receive with lst above, now its value is ['o','hello', 1]
    n='o'

So in find_letter(lst[0:]), you use list slicing, but on lst=['o','hello', 1] line, you override it again and you always execute your search on the first element of the list.

n = "o"  # you can set this too, but this is optional
def find_letter(lst):
    # do not set a value to lst in here

    if not lst:          
        return 0

    elif lst[0] == n:  # You checked first element in here
        return True

    elif find_letter(lst[1:]):  # since you checked the first element, skip it and return the orher elements of the list
        return True

    else: 
        return False

lst = ['o','hello', 1]
print find_letter(lst)

Answered By: FallenAngel

Answer 5

Since you need a recursive version:

Short version

def find_letter(let, lst):
    return (lst or False) and 
           ((isinstance(lst[0], str) and let in lst[0]) or find_letter(let, lst[1:]))

More explicit version

def find_letter(let, lst):
    if lst:
        e = lst[0]
        return (isinstance(e, str) and let in e) or find_letter(let, lst[1:])
    return False

Even more explicit version

def find_letter(let, lst):
    if lst:
        e = lst[0]
        if isinstance(e, str) and let in e:
            return True
        return find_letter(let, lst[1:])
    return False

Note that I leave out a couple of else: because they are not necessary after a return statement. If you don’t want to test for a letter in a string, but just for equality, replace let in ... by let == ....

Answered By: uselpa

Answer 6

Here is a one line lambda:

any(list(map(lambda x: True if n in str(x) else False,lst )))

data:

lst=['o','hello', 1]
n='o'

output 1:

True

note:

any() checks if any True, then True – otherwise False – so without the any() we get:

list(map(lambda x: True if n in str(x) else False,lst ))

output 2:

[True, True, False]

Answered By: Grant Shannon

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