What is the output of this puzzle?
wordlist = ["bitcoin",
crawled_text = ”’
Research produced by the University of
Cambridge estimates that in 2017,
there are 2.9 to 5.8 million unique
users using a cryptocurrency wallet,
most of them using bitcoin.
s = crawled_text.split()
res1 = True in map(lambda x: x in s, wordlist)
res2 = any(x in wordlist for x in s)
print(res1 == res2)
res2 store whether the variable
crawled_text contains a word from the
res1: The map function checks for each element x in the
wordlist whether x is an element of the split
crawled_text. The default split function divides the string along the whitespaces. The result is an iterable with three booleans, one for each word in the
Finally, we check whether one of them is True.
res2: The any function ckecks whether there is an element in the iterable that is True. As soon as it finds such a True value, this function returns True. Note that it is more efficient to use the any function to do this instead of performing a list iteration. After checking for the first word ‘bitcoin’, the function already returns True.
Hence, both variables store the same boolean value.
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