我最近使用 lambda 函数发布了一个问题,并且在回复中有人提到 lambda 已经失宠,改为使用列表推导。我对 Python 比较陌生。我进行了一个简单的测试:
import time
S=[x for x in range(1000000)]
T=[y**2 for y in range(300)]
#
#
time1 = time.time()
N=[x for x in S for y in T if x==y]
time2 = time.time()
print 'time diff [x for x in S for y in T if x==y]=', time2-time1
#print N
#
#
time1 = time.time()
N=filter(lambda x:x in S,T)
time2 = time.time()
print 'time diff filter(lambda x:x in S,T)=', time2-time1
#print N
#
#
#http://snipt.net/voyeg3r/python-intersect-lists/
time1 = time.time()
N = [val for val in S if val in T]
time2 = time.time()
print 'time diff [val for val in S if val in T]=', time2-time1
#print N
#
#
time1 = time.time()
N= list(set(S) & set(T))
time2 = time.time()
print 'time diff list(set(S) & set(T))=', time2-time1
#print N #the results will be unordered as compared to the other ways!!!
#
#
time1 = time.time()
N=[]
for x in S:
for y in T:
if x==y:
N.append(x)
time2 = time.time()
print 'time diff using traditional for loop', time2-time1
#print N
它们都打印相同的 N 所以我评论了 print stmt out (除了最后一种方式它是无序的),但是在重复测试中产生的时间差异很有趣,如这个例子所示:
time diff [x for x in S for y in T if x==y]= 54.875
time diff filter(lambda x:x in S,T)= 0.391000032425
time diff [val for val in S if val in T]= 12.6089999676
time diff list(set(S) & set(T))= 0.125
time diff using traditional for loop 54.7970001698
因此,虽然我发现列表推导总体上更易于阅读,但至少在这个示例中似乎存在一些性能问题。
所以,两个问题:
为什么 lambda 等被推到一边?
对于列表理解方式,是否有更有效的实现方式?如果不进行测试,您如何知道它更有效?我的意思是,由于额外的函数调用,lambda/map/filter 的效率应该较低,但它似乎更有效率。
保罗