replacing items in nested lists python

╄→гoц情女王★ 提交于 2019-12-01 13:28:21

What you're doing here is a shallow copy of the list, so when you change the copy, the original changes as well. What you need is deepcopy

import copy
order = [['yhjK', 'F'], 'gap', ['bcsA', 'F'], ['bcsB', 'F'], ['bcsZ', 'F'], 
'gap', ['yhjK', 'R']]
order_1 = copy.deepcopy(order)

# Now changing order_1 will not change order
order_1[1] = ['TEST LIST']
print order[1] # Prints 'gap'
print order_1[1] # Prints '['TEST LIST']
L = [['F' if x == 'R' else 'R' if x == 'F' else x for x in row] for row in order]
import copy
order_1 = copy.deepcopy(order)

Python by default only copies references to mutable values, so changing them in one place results in them being changed everywhere. Creating a deep copy means the two instances are completely independent.

order_1 = []
for item in order:
    temp = []
    for i in item:
        if i=="F":
            temp.append("R")
        elif i=="R"
            temp.append("F")
        else:
            temp.append(i)
    order_1.append(temp)
Jupiter

It seems that, as with many other languages, arrays are no more than constants (not even actual pointers although that might be different in python). Rather, The address of the first element of the array is constant. This means that if you would directly copy the array, you would only copy the address of the first element.

From your question, I understand that you want a deep copy. That means that, you also need to copy the contents of you nested arrays. You can use copy.deepcopy(x) to make a deepcopy of the array.

Check out the copy module for more in-depth information.

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