如果我想要列表中的最大值,我可以只编写max(List)
,但如果我还需要最大值的索引怎么办?
我可以这样写:
maximum=0
for i,value in enumerate(List):
if value>maximum:
maximum=value
index=i
但在我看来,这很乏味。
如果我写道:
List.index(max(List))
然后,它将对列表进行两次迭代。
有没有更好的方法?
发布于 2011-06-01 05:03:01
有很多选项,例如:
import operator
index, value = max(enumerate(my_list), key=operator.itemgetter(1))
发布于 2011-06-01 07:09:52
我认为公认的答案很好,但你为什么不明确地这样做呢?我觉得更多的人会理解你的代码,这与PEP 8是一致的:
max_value = max(my_list)
max_index = my_list.index(max_value)
这种方法也比公认的答案快三倍左右:
import random
from datetime import datetime
import operator
def explicit(l):
max_val = max(l)
max_idx = l.index(max_val)
return max_idx, max_val
def implicit(l):
max_idx, max_val = max(enumerate(l), key=operator.itemgetter(1))
return max_idx, max_val
if __name__ == "__main__":
from timeit import Timer
t = Timer("explicit(l)", "from __main__ import explicit, implicit; "
"import random; import operator;"
"l = [random.random() for _ in xrange(100)]")
print "Explicit: %.2f usec/pass" % (1000000 * t.timeit(number=100000)/100000)
t = Timer("implicit(l)", "from __main__ import explicit, implicit; "
"import random; import operator;"
"l = [random.random() for _ in xrange(100)]")
print "Implicit: %.2f usec/pass" % (1000000 * t.timeit(number=100000)/100000)
在我的电脑中运行时的结果:
Explicit: 8.07 usec/pass
Implicit: 22.86 usec/pass
其他组:
Explicit: 6.80 usec/pass
Implicit: 19.01 usec/pass
发布于 2016-01-20 07:41:04
假设列表非常大,并且假设它已经是一个np.array(),这个答案比@Escualo快33倍。我不得不减少测试次数,因为测试将查看10000000个元素,而不仅仅是100个。
import random
from datetime import datetime
import operator
import numpy as np
def explicit(l):
max_val = max(l)
max_idx = l.index(max_val)
return max_idx, max_val
def implicit(l):
max_idx, max_val = max(enumerate(l), key=operator.itemgetter(1))
return max_idx, max_val
def npmax(l):
max_idx = np.argmax(l)
max_val = l[max_idx]
return (max_idx, max_val)
if __name__ == "__main__":
from timeit import Timer
t = Timer("npmax(l)", "from __main__ import explicit, implicit, npmax; "
"import random; import operator; import numpy as np;"
"l = np.array([random.random() for _ in xrange(10000000)])")
print "Npmax: %.2f msec/pass" % (1000 * t.timeit(number=10)/10 )
t = Timer("explicit(l)", "from __main__ import explicit, implicit; "
"import random; import operator;"
"l = [random.random() for _ in xrange(10000000)]")
print "Explicit: %.2f msec/pass" % (1000 * t.timeit(number=10)/10 )
t = Timer("implicit(l)", "from __main__ import explicit, implicit; "
"import random; import operator;"
"l = [random.random() for _ in xrange(10000000)]")
print "Implicit: %.2f msec/pass" % (1000 * t.timeit(number=10)/10 )
我电脑上的结果:
Npmax: 8.78 msec/pass
Explicit: 290.01 msec/pass
Implicit: 790.27 msec/pass
https://stackoverflow.com/questions/6193498
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