Pythonic方法在列表中找到最大值及其索引？

Question

如果我想要列表中的最大值,我可以写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

但这对我来说看起来很乏味.

如果我写:

List.index(max(List))

然后它将迭代列表两次.

有没有更好的办法？

Answer 1

我认为接受的答案很棒,但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 8一致:

max_value = max(my_list)
max_index = my_list.index(max_value)

这种方法也比接受的答案快三倍:

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)

在我的电脑中运行的结果:

Explicit: 8.07 usec/pass
Implicit: 22.86 usec/pass

其他套装:

Explicit: 6.80 usec/pass
Implicit: 19.01 usec/pass

Answer 2

有很多选择,例如:

import operator
index, value = max(enumerate(my_list), key=operator.itemgetter(1))

Answer 3

这个答案比@Escualo快33倍,假设列表非常大,假设它已经是一个np.array().我不得不拒绝测试运行次数,因为测试正在查看10000000个元素而不仅仅是100个元素.

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 )

我的电脑上的结果:

Npmax: 8.78 msec/pass
Explicit: 290.01 msec/pass
Implicit: 790.27 msec/pass

Answer 4

使用Python内置库,它非常简单:

a = [2, 9, -10, 5, 18, 9] 
max(xrange(len(a)), key = lambda x: a[x])

Answer 5

max([(v,i) for i,v in enumerate(my_list)])

Answer 6

我建议一种非常简单的方法：

import numpy as np
l = [10, 22, 8, 8, 11]
print(np.argmax(l))
print(np.argmin(l))

希望能帮助到你。