两个列表元素之间的关系：如何在Python中利用它？

Question

SO here is my minimal working example:

# I have a list
list1 = [1,2,3,4]
#I do some operation on the elements of the list
list2 = [2**j for j in list1]
# Then I want to have these items all shuffled around, so for instance
list2 = np.random.permutation(list2)

#Now here is my problem: I want to understand which element of the new list2 came from which element of list1. I am looking for something like this:

list1.index(something)

# Basically given an element of list2, I want to understand from where it came from, in list1. I really cant think of a simple way of doing this, but there must be an easy way!

Can you please suggest me an easy solution? This is a minimal working example,however the main point is that I have a list, I do some operation on the elements and assign these to a new list. And then the items get all shuffled around and I need to understand where they came from.

Answer 1

You could use a permutation list/index :

# I have a list
list1 = [1,2,3,4]
#I do some operation on the elements of the list
list2 = [2**j for j in list1]
# Then I want to have these items all shuffled around, so for instance
index_list = range(len(list2))
index_list = np.random.permutation(index_list)
list3 = [list2[i] for i in index_list]

then,with input_element :

answer = index_list[list3.index(input_element)]

Answer 2

enumerate, like everyone said is the best option but there is an alternative if you know the mapping relation. You can write a function that does the opposite of the mapping relation. (eg. decodes if the original function encodes.) Then you use decoded_list = map(decode_function,encoded_list) to get a new list. Then by cross comparing this list with the original list, you can achieve your goal.

Enumerate is better if you are certain that the same list was modified using the encode_function from within the code to get the encoded list. However, if you are importing this new list from elsewhere, eg. from a table on a website, my approach is the way to go.

Answer 3

Based on your code:

# I have a list
list1 = [1,2,3,4]
#I do some operation on the elements of the list
list2 = [2**j for j in list1]
# made a recode of index and value
index_list2 = list(enumerate(list2))
# Then I want to have these items all shuffled around, so for instance
index_list3 = np.random.permutation(index_list2)
idx, list3 = zip(*index_list3)

#get the index of element_input in list3, then get the value of the index in idx, that should be the answer you want.
answer = idx[list3.index(element_input)]

Answer 4

def index3_to_1(index):
    y = list3[index]
    x = np.log(y)/np.log(2)  # inverse y=f(x) for your operation
    return list1.index(x)  

This supposes that the operations you are doing on list2 are reversible. Also, it supposes that each element in list1 is unique.