我见过多个线程有相同问题的人,但由于问题的独特性,似乎已根据具体情况提供了解决方案
这是我的代码:
loga = [(912, "Message A1") , (1000, "Message A2") , (988, "Message A3") , (1012, "Message A4") , (1002, "Message A5")]
logb = [(926, "Message B1") , (1008, "Message B2") , (996, "Message B3") , (1019, "Message B4") , (1100, "Message B5")]
logc = [(1056,"Message C1") , (1033, "Message C2") , (999, "Message C3") , (1054, "Message C4") , (1086, "Message C5")]
logs = [loga, logb, logc]
out = []
def find_lowest_i(lst):
for i in range(len(lst)):
log = lst[i]
if log:
t = log[0][0]
if i==0 or t < lowest_t:
lowest_i = i
lowest_t = t
return lowest_i
while True:
i = find_lowest_i(logs)
print "i=", i
tpl = logs[i].pop(0)
print tpl
out.append(tpl)
print out
我的确切错误:
“消息文件名行位置追溯 19 find_lowest_i 13
UnboundLocalError:分配前引用了局部变量‘lowest_t’”