blocks|key|881941|text|有很多选项，例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|881942|import+operator
index,+value+=+max(enumerate(my_list),+key=operator.itemgetter(1))|code-block|syntax|javascript|881943|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

There are many options, for example:

<pre><code>import operator
index, value = max(enumerate(my_list), key=operator.itemgetter(1))
</code></pre>

blocks|key|881968|text|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：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|881969|max_value+=+max(my_list)
max_index+=+my_list.index(max_value)|code-block|syntax|javascript|881970|这种方法也比公认的答案快三倍左右：|881971|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:+%25.2f+usec/pass"+%25+(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:+%25.2f+usec/pass"+%25+(1000000+*+t.timeit(number=100000)/100000)|881972|在我的电脑中运行时的结果：|881973|Explicit:+8.07+usec/pass
Implicit:+22.86+usec/pass|881974|其他组：|881975|Explicit:+6.80+usec/pass
Implicit:+19.01+usec/pass|881976|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

I think the accepted answer is great, but why don't you do it explicitly? I feel more people would understand your code, and that is in agreement with PEP 8:

<pre><code>max_value = max(my_list)
max_index = my_list.index(max_value)
</code></pre>

This method is also about three times faster than the accepted answer:

<pre><code>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)
</code></pre>

Results as they run in my computer:

<pre><code>Explicit: 8.07 usec/pass
Implicit: 22.86 usec/pass
</code></pre>

Other set:

<pre><code>Explicit: 6.80 usec/pass
Implicit: 19.01 usec/pass
</code></pre>

blocks|key|882009|text|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。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|882010|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:+%25.2f+msec/pass"+%25+(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:+%25.2f+msec/pass"+%25+(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:+%25.2f+msec/pass"+%25+(1000++*+t.timeit(number=10)/10+)|code-block|syntax|javascript|882011|我电脑上的结果：|882012|Npmax:+8.78+msec/pass
Explicit:+290.01+msec/pass
Implicit:+790.27+msec/pass|882013|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

This answer is 33 times faster than @Escualo assuming that the list is very large, and assuming that it's already an np.array(). I had to turn down the number of test runs because the test is looking at 10000000 elements not just 100.

<pre><code>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 )
</code></pre>

Results on my computer:

<pre><code>Npmax: 8.78 msec/pass
Explicit: 290.01 msec/pass
Implicit: 790.27 msec/pass
</code></pre>

blocks|key|882055|text|使用Python的内置库，这非常简单：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|882056|a+=+[2,+9,+-10,+5,+18,+9]+
max(xrange(len(a)),+key+=+lambda+x:+a[x])|code-block|syntax|javascript|882057|这将告诉max使用定制函数lambda+x:+a[x]在列表[0,+1,+2,+...,+len(a)]中查找最大的数字，该函数表示0实际上是2，1实际上是9，依此类推。|offset|length|style|CODE|882058|entityMap^0|0|0|4|3|D|E|U|M|1U|1|1Z|1|21|1|26|1|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]|$I|Z|J|10|K|L]|$I|11|J|12|K|L]|$I|13|J|14|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|N|$]]

With Python's built-in library, it's pretty easy:

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

This tells <code>max</code> to find the largest number in the list <code>[0, 1, 2, ..., len(a)]</code>, using the custom function <code>lambda x: a[x]</code>, which says that <code>0</code> is actually <code>2</code>, <code>1</code> is actually <code>9</code>, etc.

blocks|key|882065|text|max([(v,i)+for+i,v+in+enumerate(my_list)])|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|882066|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>max([(v,i) for i,v in enumerate(my_list)])
</code></pre>

blocks|key|882097|text|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|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|882098|如果最大值不止一次出现，且您想要获取所有索引，|unstyled|882099|max_value+=+max(your_list)
maxIndexList+=+[index+for+index,value+in+enumerate(your_list)+if+value==max(your_list)]|882100|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>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
</code></pre>

If maximum value in present more than once and you want to get all indices,

<pre><code>max_value = max(your_list)
maxIndexList = [index for index,value in enumerate(your_list) if value==max(your_list)]
</code></pre>

blocks|key|882135|text|也许你需要一个有序的列表？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|882136|试试这个：|882137|your_list+=+[13,+352,+2553,+0.5,+89,+0.4]
sorted_list+=+sorted(your_list)
index_of_higher_value+=+your_list.index(sorted_list[-1])|code-block|syntax|javascript|882138|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Maybe you need a sorted list anyway?

Try this: 

<pre><code>your_list = [13, 352, 2553, 0.5, 89, 0.4]
sorted_list = sorted(your_list)
index_of_higher_value = your_list.index(sorted_list[-1])
</code></pre>

blocks|key|665218|text|以下是使用Python的内置函数对您的问题的完整解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|665219|#+Create+the+List
numbers+=+input("Enter+the+elements+of+the+list.+Separate+each+value+with+a+comma.+Do+not+put+a+comma+at+the+end.\n").split(",")+

#+Convert+the+elements+in+the+list+(treated+as+strings)+to+integers
numberL+=+[int(element)+for+element+in+numbers]+

#+Loop+through+the+list+with+a+for-loop

for+elements+in+numberL:
++++maxEle+=+max(numberL)
++++indexMax+=+numberL.index(maxEle)

print(maxEle)
print(indexMax)|code-block|syntax|javascript|665220|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here is a complete solution to your question using Python's built-in functions:

<pre><code># Create the List
numbers = input("Enter the elements of the list. Separate each value with a comma. Do not put a comma at the end.\n").split(",") 

# Convert the elements in the list (treated as strings) to integers
numberL = [int(element) for element in numbers] 

# Loop through the list with a for-loop

for elements in numberL:
 maxEle = max(numberL)
 indexMax = numberL.index(maxEle)

print(maxEle)
print(indexMax)
</code></pre>

blocks|key|665344|text|我建议一种非常简单的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|665345|import+numpy+as+np
l+=+[10,+22,+8,+8,+11]
print(np.argmax(l))
print(np.argmin(l))|code-block|syntax|javascript|665346|希望能有所帮助。|665347|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I would suggest a very simple way:

<pre><code>import numpy as np
l = [10, 22, 8, 8, 11]
print(np.argmax(l))
print(np.argmin(l))
</code></pre>

Hope it helps.

blocks|key|665396|text|我列了一些大单子。一个是列表，另一个是数值数组。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|665397|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))|code-block|syntax|javascript|665398|使用jupyter+notebook的%25%25time函数，我可以比较各种东西的速度。|665399|2秒：|665400|%25%25time
listv.index(max(listv))|665401|54.6秒：|665402|%25%25time
listv.index(max(arrayv))|665403|6.71秒：|665404|%25%25time
np.argmax(listv)|665405|103毫秒：|665406|%25%25time
np.argmax(arrayv)|665407|numpy的数组速度非常快。|665408|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|14|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|15|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|16|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|17|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|18|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|19|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|1A|8|@]|9|@]|A|$E|F]]|$1|U|3|V|5|6|7|1B|8|@]|9|@]|A|$]]|$1|W|3|X|5|D|7|1C|8|@]|9|@]|A|$E|F]]|$1|Y|3|Z|5|6|7|1D|8|@]|9|@]|A|$]]|$1|10|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]]|11|$]]

I made some big lists. One is a list and one is a numpy array.
<pre><code>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))
</code></pre>
Using jupyter notebook's %%time function I can compare the speed of various things.
2 seconds:
<pre><code>%%time
listv.index(max(listv))
</code></pre>
54.6 seconds:
<pre><code>%%time
listv.index(max(arrayv))
</code></pre>
6.71 seconds:
<pre><code>%%time
np.argmax(listv)
</code></pre>
103 ms:
<pre><code>%%time
np.argmax(arrayv)
</code></pre>
numpy's arrays are crazy fast.

blocks|key|882713|text|列表理解方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|882714|假设您有一些列表List+=+[5,2,3,8]|offset|length|style|CODE|882715|那么[i+for+i+in+range(len(List))+if+List[i]+==+max(List)]将是一种pythonic式的列表理解方法，用于查找List[i]+==+max(List)所在位置的值"i“。|882716|对于作为列表列表的数组，它很容易扩展，只需执行for循环即可。|882717|例如，使用列表的任意列表"array“并将"index”初始化为空列表。|882718|array+=+[[5,+0,+1,+1],+
[1,+0,+1,+5],+
[0,+1,+6,+0],+
[0,+4,+3,+0],+
[5,+2,+0,+0],+
[5,+0,+1,+1],+
[0,+6,+0,+1],+
[0,+1,+0,+6]]
index+=+[]

for+List+in+array:
++++index.append([i+for+i+in+range(len(List))+if+List[i]+==+max(List)])
index|code-block|syntax|javascript|882719|输出：[[0],+[3],+[2],+[1],+[0],+[0],+[1],+[3]]|882720|entityMap^0|0|8|G|0|2|1H|28|K|0|0|0|0|3|14|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@$D|Y|E|Z|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|10|8|@$D|11|E|12|F|G]|$D|13|E|14|F|G]]|9|@]|A|$]]|$1|J|3|K|5|6|7|15|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|16|8|@]|9|@]|A|$]]|$1|N|3|O|5|P|7|17|8|@]|9|@]|A|$Q|R]]|$1|S|3|T|5|6|7|18|8|@$D|19|E|1A|F|G]]|9|@]|A|$]]|$1|U|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|V|$]]

List comprehension method:
Let's say you have some list <code>List = [5,2,3,8]</code>
Then <code>[i for i in range(len(List)) if List[i] == max(List)]</code> would be a pythonic list comprehension method to find the values &quot;i&quot; where <code>List[i] == max(List)</code>.
It is easily scalable for arrays that are lists of lists, simply by doing a for loop.
For instance, with an arbitrary list of lists &quot;array&quot; and initalizing &quot;index&quot; as an empty list.
<pre><code>array = [[5, 0, 1, 1], 
[1, 0, 1, 5], 
[0, 1, 6, 0], 
[0, 4, 3, 0], 
[5, 2, 0, 0], 
[5, 0, 1, 1], 
[0, 6, 0, 1], 
[0, 1, 0, 6]]
index = []

for List in array:
 index.append([i for i in range(len(List)) if List[i] == max(List)])
index
</code></pre>
Output: <code>[[0], [3], [2], [1], [0], [0], [1], [3]] </code>

If I want the maximum value in a list, I can just write <code>max(List)</code>, but what if I also need the index of the maximum value?

I can write something like this:

<pre><code>maximum=0
for i,value in enumerate(List):
 if value&gt;maximum:
 maximum=value
 index=i
</code></pre>

But it looks tedious to me.

And if I write:

<pre><code>List.index(max(List))
</code></pre>

Then it will iterate the list twice.

Is there a better way?

Pythonic way to find maximum value and its index in a list?

Python

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=0for i,value in enumerate(List):    if value>maximum:        maximum=value        index=i但在我看来，这很乏味...

问在列表中找到最大值及其索引的Pythonic方法？
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问在列表中找到最大值及其索引的Pythonic方法？
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