First clear solution for Three Words by werepire - python coding challenges

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First solution in Clear category for Three Words by werepire

def checkio(words):
    
    w_count=0
    w_list=words.split(' ')

    for i in range (0, len(w_list)):
        n=ord(w_list[i][0])
        if n<58 and n>47:
            w_count=0
        else:
            w_count+=1
        if(w_count>=3):
            return True
            
    return False


#These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':
    assert checkio("Hello World hello") == True, "Hello"
    assert checkio("He is 123 man") == False, "123 man"
    assert checkio("1 2 3 4") == False, "Digits"
    assert checkio("bla bla bla bla") == True, "Bla Bla"
    assert checkio("Hi") == False, "Hi"

Oct. 25, 2016

Comments:

insightfulquery on Nov. 16, 2016, 9:50 p.m.
checkout the string.isnumeric() function

DahliaSR on Nov. 16, 2016, 10:20 p.m.
Good point. But in this case I think str.isdigit() makes more sense since the the number of characters to be checked is less.

insightfulquery on Nov. 17, 2016, 4:15 p.m.
Oh? What's the difference between isnumeric and isdigit? I looked them up and they appear to be identical.

DahliaSR on Nov. 17, 2016, 7:07 p.m.
The <a href="https://docs.python.org/3/library/stdtypes.html?highlight=isnumeric#str.isdigit">python documentation</a> expalins this. + str.isdigit() checks all unicode characters woth the properties NumericType=Digit or NumericType=Decimal + str.isnumeric() checks all unicode characters woth the properties NumericType=Digit, NumericType=Decimal or Numeric_Type=Numeric So you see str.isdigit() only checks 2 character properties instead of 3.

werepire on Nov. 20, 2016, 4:31 p.m.
Thanks for your suggestions. I've rewritten the code, it's way better now.

DahliaSR on Nov. 16, 2016, 10:32 p.m.
A really bad solution. Redundant parentheses, unclear expressions and ignoring the nice features python provides.

Strings and Integers

0 %

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