Sort string with integers and words without any change in their positions

Say I have a string a.

a = "12 I have car 8 200 a"

I need to sort this string in such a way that the output should be

8 a car have 12 200 I

ie, Sort the string in such a way that all words are in alphabetical order and all integers are in numerical order. Furthermore, if the nth element in the string is an integer it must remain an integer, and if it is a word it must remain a word.

This is what I tried.

a = "12 I have car 8 200 a"


def is_digit(element_):
    """
    Function to check the item is a number. We can make using of default isdigit function
    but it will not work with negative numbers.
    :param element_:
    :return: is_digit_
    """
    try:
        int(element_)
        is_digit_ = True
    except ValueError:
        is_digit_ = False

    return is_digit_



space_separated = a.split()

integers = [int(i) for i in space_separated if is_digit(i)]
strings = [i for i in space_separated if i.isalpha()]

# sort list in place
integers.sort()
strings.sort(key=str.lower)

# This conversion to iter is to make use of next method.
int_iter = iter(integers)
st_iter = iter(strings)

final = [next(int_iter) if is_digit(element) else next(st_iter) if element.isalpha() else element for element in
         space_separated]

print " ".join(map(str, final))
# 8 a car have 12 200 I

I am getting the right output. But I am using two separate sorting function for sorting integers and the words(which I think is expensive).

Is it possible to do the entire sorting using a single sort function?.

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