loop for specific characters in a string python

Question

I got stuck on a homework problem where I need to write a function that takes in one parameter and returns True if there are exactly three 7's in a string. If there are less than three or more than three '7's, the function should return False.

def lucky_seven(a_string):
    if "7" in a_string:
       return True 
    else:
       return False

 print(lucky_seven("happy777bday"))
 print(lucky_seven("happy77bday"))
 print(lucky_seven("h7app7ybd7ay"))      

The result should be True, False, True. I got I to work where only one 7 is in a string. Appreciate if anyone can point me into the right direction.

Answer 1

you can use str.count:

str.count(sub[, start[, end]])

Return the number of non-overlapping occurrences of substring sub in the range [start, end]. Optional arguments start and end are interpreted as in slice notation.

def lucky_seven(a_string):
    if a_string.count("7") == 3:
       return True 
    else:
       return False

print(lucky_seven("happy777bday"))
print(lucky_seven("happy77bday"))
print(lucky_seven("h7app7ybd7ay"))  

Answer 2

You an do it by using regexp easily:-

import re

def lucky_seven(text):
    return len(re.findall("7", text)) is 3

print(lucky_seven("happy777bday"))
print(lucky_seven("happy77bday"))
print(lucky_seven("h7app7ybd7ay"))

Answer 3

"7" in a_string is True iff there are any 7 's in a_string ; to check for a certain number of 7 s, you could compare each character of a_string to 7 and count how many times they match. You could also do this with a regex.

Answer 4

You could just create your own counter like -

def lucky_seven(a_string):
    count = 0
    for s in a_string:
        if s == "7":
            count += 1
    return True if count == 3 else False

Or you could use python's collections module, which will return a dict of how many times an element has appeared in a list. Since a string is nothing but a list, it will work -

import collections
def lucky_seven(a_string):
    return True if collections.Counter("happy777bday")["7"] == 3 else False

Or a string counter-

def lucky_seven(a_string):
    return True if str.count(a_string, "7") == 3 else False

Or even a list comprehension solution, though really not needed -

def lucky_seven(a_string):
    return True if len([i for i in a_string if i == "7"]) == 3 else False