我建议一个非常简单的方法:

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

希望能有所帮助。

我认为公认的答案很好，但你为什么不明确地说出来呢?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:

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

其他设置:

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

max([(value,index) for index,value in enumerate(your_list)]) #if maximum value is present more than once in your list then this will return index of the last occurrence

如果最大值出现在present中不止一次，你想要得到所有的下标，

max_value = max(your_list)
maxIndexList = [index for index,value in enumerate(your_list) if value==max(your_list)]

有很多选择，例如:

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

这个答案比@Escualo快33倍，假设列表非常大，并且它已经是一个np.array()。我不得不减少测试运行的次数，因为测试要查看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

我列了一些大清单。一个是列表，一个是numpy数组。

import numpy as np
import random
arrayv=np.random.randint(0,10,(100000000,1))
listv=[]
for i in range(0,100000000):
    listv.append(random.randint(0,9))

使用jupyter笔记本的%%time功能，我可以比较各种事情的速度。

2秒:

%%time
listv.index(max(listv))

54.6秒:

%%time
listv.index(max(arrayv))

6.71秒:

%%time
np.argmax(listv)

103女士:

%%time
np.argmax(arrayv)

Numpy的数组非常快。