lambda in list comprehension acting funny

Jurko Gospodnetić jurko.gospodnetic at
Wed Jul 11 09:01:18 CEST 2012


>> funcs = [ lambda x: x**i for i in range( 5 ) ]
>> print funcs[0]( 2 )
>> This gives me
>> 16
>> When I was excepting
>> 1
>> Does anyone know why?

   Just the way Python lambda expressions bind their variable 
references. Inner 'i' references the outer scope's 'i' variable and not 
its value 'at the time the lambda got defined'.

> And more importantly, what's the simplest way to achieve the latter? :)

   Try giving the lambda a default parameter (they get calculated and 
have their value stored at the time the lambda is defined) like this:
   funcs = [ lambda x, i=i: x**i for i in range( 5 ) ]

   Hope this helps.

   Best regards,
     Jurko Gospodnetić

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