我想知道这是否可能...将列表转换为变量序列
我有一系列变量必须相应地分配给do.something(a,b)a和b变量。
事情是这样的:
# # Have a list of sequenced variables.
list = 2:90 , 1:140 , 3:-40 , 4:60
# # "Template" on where to assign the variables from the list.
do.something (a,b)
# # Assign the variables from the list in a sequence with possibility of "in between" functions like print and time.sleep() added.
do.something (2,90)
time.sleep(1)
print "Did something (%d,%d)" % (# # vars from list?)
do.something (1,140)
time.sleep(1)
print "Did something (%d,%d)" % (# # vars from list?)
do.something (3,-40)
time.sleep(1)
print "Did something (%d,%d)" % (# # vars from list?)
do.something (4,60)
time.sleep(1)
print "Did something (%d,%d)" % (# # vars from list?)
任何想法?
为什么不使用'list =((2,90),(1,140),...)'? – kennytm 2010-04-13 22:22:21
@Matti Virkkunen。 :) 谢谢你的提示。甚至不知道你必须这样做。 – wtz 2010-04-13 22:49:50