如何使用Python方式在列表中找到最大值及其索引?

2024-04-26 18:32:58 发布

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如果我想要列表中的最大值,我可以只写max(List),但是如果我还需要最大值的索引呢?

我可以这样写:

maximum=0
for i,value in enumerate(List):
    if value>maximum:
        maximum=value
        index=i

但在我看来很乏味。

如果我写:

List.index(max(List))

然后它将重复列表两次。

有更好的办法吗?


Tags: in列表forindexifvaluemaxlist
3条回答

我认为公认的答案很好,但你为什么不明确地回答呢?我觉得更多的人会理解你的代码,这与第8号政治公众人物计划一致:

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

这种方法的速度大约是公认答案的三倍:

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)

结果在我的计算机中运行时:

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

其他集合:

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

有许多选项,例如:

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

假设列表非常大,并且假设它已经是np.array(),那么这个答案比@Escualo快33倍。我不得不减少测试运行的次数,因为测试关注的是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 )

我的计算机上的结果:

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

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