Similar idea with a list comprehension but without enumerate
m = max(a)
[i for i in range(len(a)) if a[i] == m]
I came up with the following and it works as you can see with max, min and others functions over lists like these:
So, please consider the next example list find out the position of the maximum in the list a:
>>> a = [3,2,1, 4,5]
Using the generator enumerate and making a casting
>>> list(enumerate(a))
[(0, 3), (1, 2), (2, 1), (3, 4), (4, 5)]
At this point, we can extract the position of max with
>>> max(enumerate(a), key=(lambda x: x[1]))
(4, 5)
The above tells us, the maximum is in the position 4 and his value is 5.
As you see, in the key argument, you can find the maximum over any iterable object by defining a lambda appropriate.
I hope that it contributes.
PD: As @PaulOyster noted in a comment. With Python 3.x the min and max allow a new keyword default that avoid the raise exception ValueError when argument is empty list. max(enumerate(list), key=(lambda x:x[1]), default = -1)
Just one line:
idx = max(range(len(a)), key = lambda i: a[i])
a = [32, 37, 28, 30, 37, 25, 27, 24, 35,
55, 23, 31, 55, 21, 40, 18, 50,
35, 41, 49, 37, 19, 40, 41, 31]
import pandas as pd
pd.Series(a).idxmax()
9
That is how I usually do it.
If you want to get the indices of the largest n numbers in a list called data, you can use Pandas sort_values:
pd.Series(data).sort_values(ascending=False).index[0:n]
Also a solution, which gives only the first appearance, can be achieved by using numpy:
>>> import numpy as np
>>> a_np = np.array(a)
>>> np.argmax(a_np)
9
import operator
def max_positions(iterable, key=None, reverse=False):
if key is None:
def key(x):
return x
if reverse:
better = operator.lt
else:
better = operator.gt
it = enumerate(iterable)
for pos, item in it:
break
else:
raise ValueError("max_positions: empty iterable")
# note this is the same exception type raised by max([])
cur_max = key(item)
cur_pos = [pos]
for pos, item in it:
k = key(item)
if better(k, cur_max):
cur_max = k
cur_pos = [pos]
elif k == cur_max:
cur_pos.append(pos)
return cur_max, cur_pos
def min_positions(iterable, key=None, reverse=False):
return max_positions(iterable, key, not reverse)
>>> L = range(10) * 2
>>> L
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> max_positions(L)
(9, [9, 19])
>>> min_positions(L)
(0, [0, 10])
>>> max_positions(L, key=lambda x: x // 2, reverse=True)
(0, [0, 1, 10, 11])
>>> max(enumerate([1,2,3,32,1,5,7,9]),key=lambda x: x[1])
>>> (3, 32)
I can't reproduce the @SilentGhost-beating performance quoted by @martineau. Here's my effort with comparisons:
=== maxelements.py ===
a = [32, 37, 28, 30, 37, 25, 27, 24, 35, 55, 23, 31, 55, 21, 40, 18, 50,
35, 41, 49, 37, 19, 40, 41, 31]
b = range(10000)
c = range(10000 - 1, -1, -1)
d = b + c
def maxelements_s(seq): # @SilentGhost
''' Return list of position(s) of largest element '''
m = max(seq)
return [i for i, j in enumerate(seq) if j == m]
def maxelements_m(seq): # @martineau
''' Return list of position(s) of largest element '''
max_indices = []
if len(seq):
max_val = seq[0]
for i, val in ((i, val) for i, val in enumerate(seq) if val >= max_val):
if val == max_val:
max_indices.append(i)
else:
max_val = val
max_indices = [i]
return max_indices
def maxelements_j(seq): # @John Machin
''' Return list of position(s) of largest element '''
if not seq: return []
max_val = seq[0] if seq[0] >= seq[-1] else seq[-1]
max_indices = []
for i, val in enumerate(seq):
if val < max_val: continue
if val == max_val:
max_indices.append(i)
else:
max_val = val
max_indices = [i]
return max_indices
Results from a beat-up old laptop running Python 2.7 on Windows XP SP3:
>\python27\python -mtimeit -s"import maxelements as me" "me.maxelements_s(me.a)"
100000 loops, best of 3: 6.88 usec per loop
>\python27\python -mtimeit -s"import maxelements as me" "me.maxelements_m(me.a)"
100000 loops, best of 3: 11.1 usec per loop
>\python27\python -mtimeit -s"import maxelements as me" "me.maxelements_j(me.a)"
100000 loops, best of 3: 8.51 usec per loop
>\python27\python -mtimeit -s"import maxelements as me;a100=me.a*100" "me.maxelements_s(a100)"
1000 loops, best of 3: 535 usec per loop
>\python27\python -mtimeit -s"import maxelements as me;a100=me.a*100" "me.maxelements_m(a100)"
1000 loops, best of 3: 558 usec per loop
>\python27\python -mtimeit -s"import maxelements as me;a100=me.a*100" "me.maxelements_j(a100)"
1000 loops, best of 3: 489 usec per loop
Here is the max value and the indexes it appears at:
>>> from collections import defaultdict
>>> d = defaultdict(list)
>>> a = [32, 37, 28, 30, 37, 25, 27, 24, 35, 55, 23, 31, 55, 21, 40, 18, 50, 35, 41, 49, 37, 19, 40, 41, 31]
>>> for i, x in enumerate(a):
... d[x].append(i)
...
>>> k = max(d.keys())
>>> print k, d[k]
55 [9, 12]
Later: for the satisfaction of @SilentGhost
>>> from itertools import takewhile
>>> import heapq
>>>
>>> def popper(heap):
... while heap:
... yield heapq.heappop(heap)
...
>>> a = [32, 37, 28, 30, 37, 25, 27, 24, 35, 55, 23, 31, 55, 21, 40, 18, 50, 35, 41, 49, 37, 19, 40, 41, 31]
>>> h = [(-x, i) for i, x in enumerate(a)]
>>> heapq.heapify(h)
>>>
>>> largest = heapq.heappop(h)
>>> indexes = [largest[1]] + [x[1] for x in takewhile(lambda large: large[0] == largest[0], popper(h))]
>>> print -largest[0], indexes
55 [9, 12]
The chosen answer (and most others) require at least two passes through the list.
Here's a one pass solution which might be a better choice for longer lists.
Edited: To address the two deficiencies pointed out by @John Machin. For (2) I attempted to optimize the tests based on guesstimated probability of occurrence of each condition and inferences allowed from predecessors. It was a little tricky figuring out the proper initialization values for max_val and max_indices which worked for all possible cases, especially if the max happened to be the first value in the list — but I believe it now does.
def maxelements(seq):
''' Return list of position(s) of largest element '''
max_indices = []
if seq:
max_val = seq[0]
for i,val in ((i,val) for i,val in enumerate(seq) if val >= max_val):
if val == max_val:
max_indices.append(i)
else:
max_val = val
max_indices = [i]
return max_indices
You can do it in various ways.
The old conventional way is,
maxIndexList = list() #this list will store indices of maximum values
maximumValue = max(a) #get maximum value of the list
length = len(a) #calculate length of the array
for i in range(length): #loop through 0 to length-1 (because, 0 based indexing)
if a[i]==maximumValue: #if any value of list a is equal to maximum value then store its index to maxIndexList
maxIndexList.append(i)
print(maxIndexList) #finally print the list
Another way without calculating the length of the list and storing maximum value to any variable,
maxIndexList = list()
index = 0 #variable to store index
for i in a: #iterate through the list (actually iterating through the value of list, not index )
if i==max(a): #max(a) returns a maximum value of list.
maxIndexList.append(index) #store the index of maximum value
index = index+1 #increment the index
print(maxIndexList)
We can do it in Pythonic and smart way! Using list comprehension just in one line,
maxIndexList = [i for i,j in enumerate(a) if j==max(a)] #here,i=index and j = value of that index
All my codes are in Python 3.
This code is not as sophisticated as the answers posted earlier but it will work:
m = max(a)
n = 0 # frequency of max (a)
for number in a :
if number == m :
n = n + 1
ilist = [None] * n # a list containing index values of maximum number in list a.
ilistindex = 0
aindex = 0 # required index value.
for number in a :
if number == m :
ilist[ilistindex] = aindex
ilistindex = ilistindex + 1
aindex = aindex + 1
print ilist
ilist in the above code would contain all the positions of the maximum number in the list.
You can also use the numpy package:
import numpy as np
A = np.array(a)
maximum_indices = np.where(A==max(a))
This will return an numpy array of all the indices that contain the max value
if you want to turn this to a list:
maximum_indices_list = maximum_indices.tolist()
@shash answered this elsewhere
A Pythonic way to find the index of the maximum list element would be
position = max(enumerate(a), key=lambda x: x[1])[0]
Which does one pass. Yet, it is slower than the solution by @Silent_Ghost and, even more so, @nmichaels:
for i in s m j n; do echo $i; python -mtimeit -s"import maxelements as me" "me.maxelements_${i}(me.a)"; done
s
100000 loops, best of 3: 3.13 usec per loop
m
100000 loops, best of 3: 4.99 usec per loop
j
100000 loops, best of 3: 3.71 usec per loop
n
1000000 loops, best of 3: 1.31 usec per loop
a.index(max(a))
will tell you the index of the first instance of the largest valued element of list a.
Source: Stackoverflow.com