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I have the following code:
s = [1,2,3]
t = reversed(s)
for i in t:
print(i)
# output: 3,2,1
If I pop one element from s (original), then the t (reversed) is emptied:
s = [1,2,3]
t = reversed(s)
s.pop()
for i in t:
print(i)
# expected output: 2, 1
# actual output (nothing):
Why does this happen?
961
pop method removes the item at the given position in the list and returns it.
One more way to reverse the list in python is by using two built-in functions append() and pop(). This works by poping the last element from the original list and appending it to the new list.
In order to reverse a list without using the built-in reverse() function, we use the Slicing Operator. The slicing operator is another method used for reversing the data elements.
If you want to create a reversed copy of an existing list in Python, then you can use reversed (). With a list as an argument, reversed () returns an iterator that yields items in reverse order: In this example, you call reversed () with digits as an argument. Then you store the resulting iterator in reversed_digits.
As you already know, the call to list () consumes the iterator that results from calling reversed (). This way, you create a new list as a reversed copy of the original one. Python 2.4 added reversed (), a universal tool to facilitate reverse iteration over sequences, as stated in PEP 322.
Method 2: Using the reverse () built-in function. Using the reverse () method we can reverse the contents of the list object in-place i.e., we don’t need to create a new list instead we just copy the existing elements to the original list in reverse order. This method directly modifies the original list.
Note that .reverse () doesn’t return a new list but None: Trying to assign the return value of .reverse () to a variable is a common mistake related to using this method. The intent of returning None is to remind its users that .reverse () operates by side effect, changing the underlying list.
Taking a look at the cpython code on GitHub, we can get some intuition as to why it no longer works.
The iterator that is returned essentially requires knowing the position of the last index and the length of the array. If the size of the array is changed, the iterator will no longer work.
This will not produce the correct results either, but the iterator does run:
s = [1,2,3] t = reversed(s) s.append(4) for i in t: print(i) # output: [3, 2, 1] s = [1,2,3] t = reversed(s) s.pop() s.append(4) for i in t: print(i) # output: [4, 2, 1] It still works!
So there's an internal check to see whether or not the last index is still valid, and if it is, it's a simple for loop down to index 0.
If it doesn't work, the iterator returns empty.
161
calling reversed return a iterator over that list, which a special object that allow you iterate in reverse order over the original list, is not a new list and is a one time use only
>>> s= [1,2,3] >>> t = reversed(s) >>> t <list_reverseiterator object at 0x00000261BE8F0C40> >>> list(t) [3, 2, 1] >>> list(t) [] >>> and because this iterator reference the original list, any change on it is reflected when you iterate over the iterator later.
Update
In particular and as MZ explain, if that change is such that the state of the list is different from when the iterator was created you get nothing if the size decreases or an incomplete version of the list if increased
>>> s= [1,2,3] >>> t = reversed(s) >>> s.insert(0,23) >>> s [23, 1, 2, 3] >>> list(t) [2, 1, 23] >>> t = reversed(s) >>> s.append(32) >>> list(t) [3, 2, 1, 23] >>> s [23, 1, 2, 3, 32] >>> t = reversed(s) >>> s.pop() 32 >>> list(t) [] >>> If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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