Train creative solution for Three Words by veky - python coding challenges

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Train solution in Creative category for Three Words by veky

def checkio(words):
    d = words.split()
    return any(''.join(values).isalpha() for values in zip(d, d[1:], d[2:]))

Sept. 10, 2016

Comments:

khblackpearl on Aug. 18, 2018, 8:54 a.m.
One question about zip() Why it was zip(d, d[1:], d[2:])) instead of zip(d)?

veky on Aug. 20, 2018, 10 a.m.
zip(d) makes not much sense. It would just give you onetuples. Maybe you meant zip(*d), but here it would give you tuples of letters, while I want words I can join (so I can control how many of them I take). In essence, I want to check the following joins: <pre class='brush: python'>d[0] + d[1] + d[2] d[1] + d[2] + d[3] d[2] + d[3] + d[4] d[3] + d[4] + d[5] d[5] + d[6] + d[7] ... </pre> and if you look at the columns, you'll see `d`, `d[1:]` and `d[2:]` there. So I need just to transpose that matrix, and zip is perfect for that. Any more questions?

khblackpearl on Aug. 20, 2018, 1:55 p.m.
According to the explanation, below is my understanding. Hopefully it's the same as what you were referring to. Example: words = "Hello World hello" d = words.split() zip(d, d[1:], d[2:]) returns ('Hello', 'World', 'hello') zip(d) returns ('Hello',) ('World',) ('hello',) zip(*d) returns (not sure what * means) ('H', 'W', 'h') ('e', 'o', 'e') ('l', 'r', 'l') ('l', 'l', 'l') ('o', 'd', 'o')

veky on Aug. 20, 2018, 3:55 p.m.
Why haven't you run it? :-) And, preferably, on a Python, not a language that will become unsupported in 16 months? Besides, I really don't think you can base your understanding on a single example, especially when that example has exactly three words. Try with more words and try to predict the outcome, then you'll see whether you understand. <hr /> BTW, surely you've used * in function definitions before? <pre class='brush: python'>def f(*args): ... </pre> It means the same in a function call: if t = [a, b, c], then f(t) is f([a, b, c]), but f(*t) is f(a, b, c). Right?

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