比较和乘以列表中的元素

Question

I'm trying to write in an algorithm a function that:

• Check if all elements in a list are different
• Multiply all elements in the list, except the zeros

But I can't find a way to compare all elements in one list, do you have any idea ?

Thanks!

PS: I use arr = np.random.randint(10, size=a) to create a random list

EDIT:

More precisely, I'm trying to check if, in a numpy array to be more precise, all the elements are the same or different, if they are different, that it returns me True. Also, once that done, multiply all elements in the array except the zeros For example: If I have an array [4,2,6,8,9,0] , the algorithm tells returns me at first True because all elements are different, then it multiplies them 4*2*6*8*9 except the 0

Answer 1

You can use numpy.unique .

Following code snippet checks if all elements in the array are unique (different from each other) and if so, it will multiply non-zero values with factor factor :

import numpy as np

factor = 5

if np.unique(arr).size == arr.size:
    arr[arr != 0] = arr[arr != 0] * factor

Answer 2

To check if all elements in a list are different you can convert the list into a set which removes duplicates and compare the length of the set to the original list. If the length of the set is different than the length of the list, then there are duplicates.

x = np.random.randint(10, size=10)
len(set(x)) == len(x)

To multiply all values except 0 you can do list comprehension to remove the 0s and use np.prod to multiply the values in the new list.

np.prod([i for i in x if i != 0])

Example:

x = np.random.randint(10, size=10)
if len(set(x)) == len(x):
    print(np.prod([i for i in x if i != 0]))
else:
    print("array is not unique")

Answer 3

You can use Collections to find the unique numbers. I have included a code that solves your problem.

import numpy as np
from collections import Counter


a = 5
arr = np.random.randint(10, size=a)


result = 1 #Variable that will store the product
flag = 0 #The counter

#Check if all the numbers are unique
for i in Counter(arr):
    if Counter(arr)[i] > 1:
        flag = 1
        break

#Convert the dictionary into a list
l = [i for i in Counter(arr)]

#Return the product of all the numbers in the list except 0
if flag == 0:
    for i in l:
        if i != 0:
            result = result * i
else:
    print("The numbers are not unique")

Answer 4

Just for fun, here's a one-liner:

arr = np.array([1, 2, 3, 4, 0])

np.prod(arr[arr!=0]) if np.unique(arr).size == arr.size else False

>>> 24

If the array is [1, 2, 3, 4, 4] the result is False