CODEWITHSUNDEEP
pythonpython-3.xlistindexing
Vesper
Vesper

Reputation: 845

How to get index of the max element from two lists after sorting it?

I have three lists price1, price2 and deviation. I wanted to find the top three highest price between price1 and price2. So to achieve that I first sorted them in decreasing order then I took the first 3 elements from both the list price1 and price2, and after that, I found the max between each max. Now I want to know the original position of the max value that is obtained after all this so that I can put it in deviation. Before sorting price1 and price2:

price1 = [12,1,2,4,45,67,433,111,45,12] #Total 10 elements in all the lists
price2 = [23,12,233,11,54,232,323,12,42,4]
deviation = [23,45,56,67,78,786,45,34,2,1]

After sorting to get the top 3 prices:

print("Price1: ",sorted(price1, reverse=True))
print("Price2: ",sorted(price2, reverse=True))

output:

Price1:  [433, 111, 67]
Price2:  [323, 233, 232]

To get the max from it:

sortedPrice1 = sorted(price1, reverse=True)
sortedPrice2 = sorted(price2, reverse=True)
print("Price1: ", sortedPrice1[:3])
print("Price2: ", sortedPrice2[:3])
for i,j in zip(sortedPrice1[:3], sortedPrice2[:3]):
    print("max: ", max(i, j))

output:

Price1:  [433, 111, 67]
Price2:  [323, 233, 232]
max:  433
max:  233
max:  232

Now What I want is that I want to find the postions of these max values. for example:

433 is in `price1` at 6th position
233 is in `price2` at 2nd position
232 is in `price2` at 5th position

and ultimately in the end I want to put these positions into the deviation list to get the value in front of these prices. so:

deviation[6] = 45
deviations[2] = 56
deviations[5] = 786

Upvotes: 1

Views: 266

Answers (5)

Andrej Kesely
Andrej Kesely

Reputation: 195418

No need to use list.index(), zip() is enough:

price1 = [12,1,2,4,45,67,433,111,45,12] #Total 10 elements in all the lists
price2 = [23,12,233,11,54,232,323,12,42,4]
deviation = [23,45,56,67,78,786,45,34,2,1]

for a, b, _ in zip( sorted(zip(price1, deviation), key=lambda k: k[0], reverse=True),
                    sorted(zip(price2, deviation), key=lambda k: k[0], reverse=True), range(3) ):  # range(3) because we want max 3 elements
    print(max(a, b, key=lambda k: k[0]))

Prints:

(433, 45)
(233, 56)
(232, 786)

EDIT: To have sublists in price1/price2/deviation lists, you can do:

price1 = [[12, 2, 3, 4],[1, 2, 5, 56],[12,34,45,3],[23,2,3,4],[1,6,55,34]]
price2 = [[1, 2, 3, 4],[1, 2, 3, 4],[1, 2, 3, 4],[1, 2, 3, 4],[1, 2, 3, 4]]
deviation = [[10, 20, 30, 40],[10, 20, 30, 40],[10, 20, 30, 40],[10, 20, 30, 40],[10, 20, 30, 40]]

for p1, p2, dev in zip(price1, price2, deviation):
    print('price1=', p1)
    print('price2=', p2)
    print('dev=', dev)
    for a, b, _ in zip( sorted(zip(p1, dev), key=lambda k: k[0], reverse=True),
                        sorted(zip(p2, dev), key=lambda k: k[0], reverse=True), range(3) ):  # range(3) because we want max 3 elements
        print(max(a, b, key=lambda k: k[0]))
    print()

Prints:

price1= [12, 2, 3, 4]
price2= [1, 2, 3, 4]
dev= [10, 20, 30, 40]
(12, 10)
(4, 40)
(3, 30)

price1= [1, 2, 5, 56]
price2= [1, 2, 3, 4]
dev= [10, 20, 30, 40]
(56, 40)
(5, 30)
(2, 20)

price1= [12, 34, 45, 3]
price2= [1, 2, 3, 4]
dev= [10, 20, 30, 40]
(45, 30)
(34, 20)
(12, 10)

price1= [23, 2, 3, 4]
price2= [1, 2, 3, 4]
dev= [10, 20, 30, 40]
(23, 10)
(4, 40)
(3, 30)

price1= [1, 6, 55, 34]
price2= [1, 2, 3, 4]
dev= [10, 20, 30, 40]
(55, 30)
(34, 40)
(6, 20)

Upvotes: 1

AfiJaabb
AfiJaabb

Reputation: 306

You can do it it two lines of code.

First find the index.

item_index_in_original_list = price1.index(433)

Then use it in your deviation list

deviation_of_max_item = deviation[item_index_in_original_list]

To repeat this for other items you can make a function with these two lines and call it.

def find_deviation(item_value,original_list):
    item_index_in_original_list = original_list.index(item_value)
    return deviation[item_index_in_original_list]

required_deviation = find_deviation(433, price1)

Upvotes: 1

Prathamesh
Prathamesh

Reputation: 1057

First create a copy of all 3 lists such as,

price1 = [12,1,2,4,45,67,433,111,45,12] #Total 10 elements in all the lists
price2 = [23,12,233,11,54,232,323,12,42,4]
deviation = [23,45,56,67,78,786,45,34,2,1]

d_p1=price1
d_p2=price2
d_dev=deviation

then perform all the operation to get max elements on this duplicate lists, and then when you have max elements in a list such as,

max_l=[433,233,232]

traverse through this list to get position,

for i  in max_l:
   if i in price_1:
       print("index:",price_1.index(i))
   else:
       print("index:",price_2.index(i))

if the positions of elements are changing after performing some operations then this process can be helpful to find original positions.

Upvotes: 1

Celius Stingher
Celius Stingher

Reputation: 18367

You can do it with a try / except too but with the same limitations as @Yberman states in his answer but without needing to define any other variable than the lists itself:

price1 = [12,1,2,4,45,67,433,111,45,12] #Total 10 elements in all the lists
price2 = [23,12,233,11,54,232,323,12,42,4]
deviation = [23,45,56,67,78,786,45,34,2,1]
for i,j in zip(sorted(price1,reverse=True)[:3], sorted(price2,reverse=True)[:3]):
    try:
      print(deviation[price1.index(max(i, j))])
    except ValueError:
      print(deviation[price2.index(max(i,j))])

Output:

45
56
786

Upvotes: 1

yberman
yberman

Reputation: 318

The list method index returns the position of an element in a list. If you replace your max function with an if statement, you can produce the desired output

for i,j in zip(sortedPrice1[:3], sortedPrice2[:3]):
    if i > j:
        print("%d is in `price1` at the %dth position" % (i, price1.index(i))
    else:
        print("%d is in `price2` at the %dth position" % (j, price2.index(j))

Note that this code:

  1. Doesn't deal with the case i=j
  2. Doesn't deal with multiple members of price1/price2 which are both in the top three.

Upvotes: 1

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