Khalid Al-Ghamdi wrote:
Hi everyone!
I'm using python 3.1 and I want to to know why is it when I enter the
following in a dictionary comprehension:
dc={y:x for y in list("khalid") for x in range(6)}
I get the following:
{'a': 5, 'd': 5, 'i': 5, 'h': 5, 'k': 5, 'l': 5}
instead of the expected:
{'a': 0, 'd': 1, 'i': 2, 'h': 3, 'k': 4, 'l': 5}
and is there a way to get the target (expected) dictionary using a
dictionary comprehension.
note that I tried sorted(range(6)) also but to no avail.
thanks
You're confused about what two for loops do here. It's basically a
doubly-nested loop, with the outer loop iterating from k through d, and
the inner loop iterating from 0 to 5. So there are 36 entries in the
dictionary, but of course the dictionary overwrites all the ones with
the same key. For each letter in the outer loop, it iterates through
all six integers, and settles on 5.
To do what you presumably want, instead of a doubly nested loop you need
a single loop with a two-tuple for each item, consisting of one letter
and one digit.
dc = { y:x for y,x in zip("khalid", range(6)) }
The output for this is:
{'a': 2, 'd': 5, 'i': 4, 'h': 1, 'k': 0, 'l': 3}
Now, this isn't the same values for each letter as you "expected," but
I'm not sure how you came up with that particular order.I expect, and
get, 0 for the first letter 'k' and 1 for the 'h'. etc.
Perhaps printing out the zip would make it clearer:
list( zip("khalid", range(6)) )
yields
[('k', 0), ('h', 1), ('a', 2), ('l', 3), ('i', 4), ('d', 5)]
DaveA
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