On 10/8/17 1:00 AM, Andrew Z wrote:
and how about adding "IF" into the mix ?
as in :
a=0
dict= {10: ['a',1,'c'], 20: ['d',2,'f']}
for i in dict:
p+= 10 if dict[i][1] in [1,2,3,4,5] else 0
can i "squish" "for" and "if" together ? or will it be too perl-ish ?
sum(10 for v in dict.values() if v[1] in [1,2,3,4,5])
or:
sum((10 if v[1] in [1,2,3,4,5] else 0) for v in dict.values())
(in case you actually need something other than 0.)
Also, note that if your list of five values is actually much longer than
five, then you want a set:
sum((10 if v[1] in {1,2,3,4,5} else 0) for v in dict.values())
--Ned.
On Sun, Oct 8, 2017 at 12:37 AM, Andrew Z <form...@gmail.com> wrote:
Nathan, Bob - on the money. Thank you !
On Sat, Oct 7, 2017 at 11:30 PM, bob gailer <bgai...@gmail.com> wrote:
On 10/7/2017 11:17 PM, Nathan Hilterbrand wrote:
dict= {10: ['a',1,'c'], 20: ['d',2,'f']}
p = sum([dict[i][1] for i in dict])
Something like that?
Ah, but that's 2 lines.
sum(val[1] for val in {10: ['a',1,'c'], 20: ['d',2,'f']}.values())
On Sat, Oct 7, 2017 at 11:07 PM, Andrew Z <form...@gmail.com> wrote:
Hello,
i wonder how can i accomplish the following as a one liner:
dict= {10: ['a',1,'c'], 20: ['d',2,'f']}
p = 0
for i in dict:
p += dict[i][1]
Thank you
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