Alister於 2012年7月12日星期四UTC+8下午5時44分15秒寫道： > On Wed, 11 Jul 2012 08:43:11 +0200, Daniel Fetchinson wrote: > > >> funcs = [ lambda x: x**i for i in range( 5 ) ] > >> print funcs[0]( 2 ) > >> print funcs[1]( 2 ) > >> print funcs[2]( 2 ) > >> > >> This gives me > >> > >> 16 16 16 > >> > >> When I was excepting > >> > >> 1 > >> 2 > >> 4 > >> > >> Does anyone know why? > > > > And more importantly, what's the simplest way to achieve the latter? > :) > > Having read Steve's explanation in the other thread (which I think has > finally flipped the light switch on lambda for me) it only requires a > minor change > > funcs=[ lambda x,y=i:x**y for i in range(5) ] > > although I cant actually think why this construct would be needed in > practice, how are you actually using it > > > -- > * Simunye is so happy she has her mothers gene's > <Dellaran> you better give them back before she misses them!
Uhn, there are 5 objects in the list if not factored well to be executed in the run time. -- http://mail.python.org/mailman/listinfo/python-list
