Python：在python中找到两个数组之间的差异

Question

I have two arrays in python.

arr1 = [(7, 0.78, 7920), (8, 0.9, 9000)]
arr2 = [(7, 1.68, 8460)]

In this array first value is ID (7,8,7).I want result basis on ID . If ID same then get difference between others two value.
Between arr1 and arr2 ID 7 is same then will be subtraction (1.68-0.78 = 0.9) And (8460-7920 = 540) otherwise it will be same. How to get Result like that ?

diffArray = [(7, 0.9, 540), (8, 0.9, 9000)]

Answer 1

Here's one way of doing it. First, I'm creating dictionaries from ID to the other values:

arr1 = [(7, 0.78, 7920), (8, 0.9, 9000)]
arr2 = [(7, 1.68, 8460)]

dict1 = {i: (x, y) for i, x, y in arr1}
dict2 = {i: (x, y) for i, x, y in arr2}

Then I'm using a list comprehension with a condition to do the right thing:

diffArray = [
    (
        i,
        abs(dict2[i][0] - dict1[i][0]),
        abs(dict2[i][1] - dict1[i][1]),
    )
    if i in dict2
    else (i, dict1[i][0], dict1[i][1])
    for i in dict1
]

The result is maybe a bit surprising:

[(7, 0.8999999999999999, 540), (8, 0.9, 9000)]

the apparent imprecision is due to the floating point representation. To get around that (if it is important), use Decimal or round the values.

Answer 2

This will go through every element in arr2 then add it into the correct index of arr1. If it doesnt exist then it will add it in fresh

arr1 = [(7, 0.78, 7920), (8, 0.9, 9000)]
arr2 = [(7, 1.68, 8460), (6,1,1)]

for j in arr2:
    for ix, i in enumerate(arr1):
        if i[0] == j[0]:
            arr1[ix] = (i[0], j[1]-i[1], j[2]-i[2])
            break
    else:
        arr1.append(j)
arr1

[(7, 0.8999999999999999, 540), (8, 0.9, 9000), (7, 1.68, 8460), (6, 1, 1)]

Answer 3

arr1 = [(7, 0.78, 7920), (8, 0.9, 9000)]
arr2 = [(7, 1.68, 8460)]
arr1.sort(key=lambda x:x[0])
arr2.sort(key=lambda x:x[0])
from itertools import zip_longest
new_list = []
for value1, value2 in zip_longest(arr1, arr2):
    if value2:
        _id1, val3, val4 = value1
        _id2, val5, val6 = value2
        if _id1 == _id2:
            diff1 = val5 - val3
            diff2 = val6 - val4
            new_list.append((_id1, diff1, diff2))
    else:
        new_list.append(value1)
print(new_list)
>>>[(7, 0.8999999999999999, 540), (8, 0.9, 9000)]

Answer 4

Here is my attempt, using two list comprehensions:

arr1 = [(7, 0.78, 7920), (8, 0.9, 9000)]
arr2 = [(7, 1.68, 8460)]

common = [tuple([a[0], b[1] - a[1], b[2] - a[2]]) for a in arr1 for b in arr2 if a[0]==b[0]]
others = [a for a in arr1 for b in arr2 if a[0]!=b[0]]
result = common + others

Output:

[(7, 0.8999999999999999, 540), (8, 0.9, 9000)]

Answer 5

Here's one way of doing it.

def diffArrays(arr1, arr2):

"""
Parameters
----------
arr1 : array
    firt array.
arr2 : array
    second array to be compared with.

Returns
-------
arr3: array
    array containing only the different elements.
"""
arr3 = []
[arr3.append(el) if el not in arr2 else '' for el in arr1]
[arr3.append(el) if el not in arr1 else '' for el in arr2]

return arr3