Develop Reference

Function which removes the first item in a list (Python)

I am trying to write a function which removes the first item in a Python list. This is what I've tried. Why doesn't remove_first_wrong change l when I call the function on it? And why does the list slicing approach work when I do it in the main function?
def remove_first_wrong(lst):
lst = lst[1:]
def remove_first_right(lst):
lst.pop(0)
if __name__ == '__main__':
l = [1, 2, 3, 4, 5]
remove_first_wrong(l)
print(l)
l_2 = [1, 2, 3, 4, 5]
remove_first_right(l_2)
print(l_2)
# Why does this work and remove_first_wrong doesn't?
l_3 = [1, 2, 3, 4, 5]
l_3 = l_3[1:]
print(l_3)

Slicing a list returns a new list object, which is a copy of the original list indices you indicated in the slice. You then rebound lst (a local name in the function) to reference that new list instead. The old list is never altered in that process.
list.pop() on the other hand, operates on the list object itself. It doesn't matter what reference you used to reach the list.
You'd see the same thing without functions:
>>> a = [1, 2]
>>> b = a[:] # slice with all the elements, produces a *copy*
>>> b
[1, 2]
>>> a.pop() # remove an element from a won't change b
2
>>> b
[1, 2]
>>> a
[1]
Using [:] is one of two ways of making a shallow copy of a list, see How to clone or copy a list?
You may want to read or watch Ned Batchelder's Names and Values presestation, to further help understand how Python names and objects work.

Inside the function remove_first_wrong the = sign reassigns the name lst to the object on the right. Which is a brand new object, created by slicing operation lst[1:]. Thus, the object lst assigned to is local to that function (and it actually will disappear on return).
That is what Martijn means by "You then rebound lst (a local name in the function) to reference that new list instead."
On contrary, lst.pop(0) is a call to the given object -- it operates on the object.
For example, this will work right too:
def remove_first_right2(lst):
x = lst # x is assigned to the same object as lst
x.pop(0) # pop the item from the object

Alternately, you can use del keyword:
def remove_first_element(lst):
del lst[0]
return lst

Append to a list defined in a tuple - is it a bug? [duplicate]

This question already has answers here:
a mutable type inside an immutable container
(3 answers)
Closed 6 years ago.
So I have this code:
tup = ([1,2,3],[7,8,9])
tup[0] += (4,5,6)
which generates this error:
TypeError: 'tuple' object does not support item assignment
While this code:
tup = ([1,2,3],[7,8,9])
try:
tup[0] += (4,5,6)
except TypeError:
print tup
prints this:
([1, 2, 3, 4, 5, 6], [7, 8, 9])
Is this behavior expected?
Note
I realize this is not a very common use case. However, while the error is expected, I did not expect the list change.

Yes it's expected.
A tuple cannot be changed. A tuple, like a list, is a structure that points to other objects. It doesn't care about what those objects are. They could be strings, numbers, tuples, lists, or other objects.
So doing anything to one of the objects contained in the tuple, including appending to that object if it's a list, isn't relevant to the semantics of the tuple.
(Imagine if you wrote a class that had methods on it that cause its internal state to change. You wouldn't expect it to be impossible to call those methods on an object based on where it's stored).
Or another example:
>>> l1 = [1, 2, 3]
>>> l2 = [4, 5, 6]
>>> t = (l1, l2)
>>> l3 = [l1, l2]
>>> l3[1].append(7)
Two mutable lists referenced by a list and by a tuple. Should I be able to do the last line (answer: yes). If you think the answer's no, why not? Should t change the semantics of l3 (answer: no).
If you want an immutable object of sequential structures, it should be tuples all the way down.
Why does it error?
This example uses the infix operator:
Many operations have an “in-place” version. The following functions
provide a more primitive access to in-place operators than the usual
syntax does; for example, the statement x += y is equivalent to x =
operator.iadd(x, y). Another way to put it is to say that z =
operator.iadd(x, y) is equivalent to the compound statement z = x; z
+= y.
https://docs.python.org/2/library/operator.html
So this:
l = [1, 2, 3]
tup = (l,)
tup[0] += (4,5,6)
is equivalent to this:
l = [1, 2, 3]
tup = (l,)
x = tup[0]
x = x.__iadd__([4, 5, 6]) # like extend, but returns x instead of None
tup[0] = x
The __iadd__ line succeeds, and modifies the first list. So the list has been changed. The __iadd__ call returns the mutated list.
The second line tries to assign the list back to the tuple, and this fails.
So, at the end of the program, the list has been extended but the second part of the += operation failed. For the specifics, see this question.

Well I guess tup[0] += (4, 5, 6) is translated to:
tup[0] = tup[0].__iadd__((4,5,6))
tup[0].__iadd__((4,5,6)) is executed normally changing the list in the first element. But the assignment fails since tuples are immutables.

Tuples cannot be changed directly, correct. Yet, you may change a tuple's element by reference. Like:
>>> tup = ([1,2,3],[7,8,9])
>>> l = tup[0]
>>> l += (4,5,6)
>>> tup
([1, 2, 3, 4, 5, 6], [7, 8, 9])

The Python developers wrote an official explanation about why it happens here: https://docs.python.org/2/faq/programming.html#why-does-a-tuple-i-item-raise-an-exception-when-the-addition-works
The short version is that += actually does two things, one right after the other:
Run the thing on the right.
assign the result to the variable on the left
In this case, step 1 works because you’re allowed to add stuff to lists (they’re mutable), but step 2 fails because you can’t put stuff into tuples after creating them (tuples are immutable).
In a real program, I would suggest you don't do a try-except clause, because tup[0].extend([4,5,6]) does the exact same thing.

Remove matching items in a list

Sorry if this is a duplicate question, I searched and couldn't find anything to help.
I'm currently trying to compare two lists. If there are any matching items I will remove them all from one of the lists.
However the results I have are buggy. Here is a rough but accurate representation of the method I'm using:
>>> i = [1,2,3,4,5,6,7,8,9]
>>> a = i
>>> c = a
>>> for b in c:
if b in i:
a.remove(b)
>>> a
[2, 4, 6, 8]
>>> c
[2, 4, 6, 8]
So I realised that the main issue is that as I remove items it shortens the list, so Python then skips over the intermediate item (seriously annoying). As a result I made a third list to act as an intermediate that can be looped over.
What really baffles me is that this list seems to change also even when I haven't directly asked it to!

In python, when you write this:
i = [1,2,3,4,5,6,7,8,9]
You create an Object (in this case, a list) and you assign it to the name i. Your next line, a = i, tells the interpreter that the name a refers to the same Object. If you want them to be separate Object you need to copy the original list. You can do that via the slicing shorthand, i[:], or you can use a = list(i) to be more explicit.

The easiest way to do this is use a set to determine shared items in a and b:
for x in set(a).intersection(b):
a.remove(x)

Your statements a = i and c = a merely make new names that reference the same object. Then as you removed things from a, it's removed from b and i, since they are the same object. You'll want to make copies of the lists instead, like so
a = i[:]
c = a[:]

a = i Doesn't make a copy of a list, it just sets another variable, i to point at your list a. Try something like this:
>>> i = [1, 2, 3, 2, 5, 6]
>>> s = []
>>> for i in t:
if i not in s:
s.append(i)
>>> s
[1, 2, 3, 5, 6]
You can also use set which guarantees no duplicates, but doesn't preserve the order:
list(set(i))

How is the iadd operator ( += ) different than simple addition? [duplicate]

This question already has answers here:
When is "i += x" different from "i = i + x" in Python?
(3 answers)
Closed 9 years ago.
I always thought x += 1 was just syntactic shorthand (and exactly equivalent to) x = x + 1, until I spent a while trying to figure out why this code wasn't acting as intended:
[ipython/euler 72 ]$ def func(mylist):
mylist += random.sample(range(100),2)
# do stuff with the random result, then restore original list
mylist = mylist[:-2]
It's supposed to return the same list it gets, but it doesn't seem to work that way:
[ipython/euler 81 ]$ x = [1,2,3]
[ipython/euler 82 ]$ func(x)
[1, 2, 3, 23, 7]
[ipython/euler 83 ]$ func(x)
[1, 2, 3, 23, 7, 42, 36]
[ipython/euler 84 ]$ func(x)
[1, 2, 3, 23, 7, 42, 36, 0, 5]
If I change the assignment statement to the long form mylist = mylist + ..., it works as expected and leaves the list unchanged.
Why is this happening? I assume it has to do with lists being mutable and possibly iadd not being 'real' addition when called as an overloaded method of list, but I still expected the interpreter to see them as equivalent.

The first code (mylist += random.sample(range(100),2)) modifies the list in place, as you've realised.
Both of these:
mylist = mylist + ...
mylist = mylist[:-2]
create a new object called mylist, which is then given the same value as (or one derived from) the previous mylist.
The latter line does not restore the original list for this reason; it's creating a new copy, and leaving the original reference alone.
You can get your desired behaviour by changing the list in place:
mylist[:] = mylist[:-2]

The line
mylist += random.sample(range(100),2)
mutates the list mylist in-place (that's why it's called iadd: in-place add). This means that it changes the original list in the caller's scope.
mylist = mylist[:-2]
creates a new local object mylist and assigns the contents of global mylist[:-2] to it. This new local object is then discarded immediately upon returning from the function.

Why does "return list.sort()" return None, not the list? [duplicate]

This question already has answers here:
Why do these list operations (methods: clear / extend / reverse / append / sort / remove) return None, rather than the resulting list?
(6 answers)
Closed 7 months ago.
I've been able to verify that the findUniqueWords does result in a sorted list. However, it does not return the list. Why?
def findUniqueWords(theList):
newList = []
words = []
# Read a line at a time
for item in theList:
# Remove any punctuation from the line
cleaned = cleanUp(item)
# Split the line into separate words
words = cleaned.split()
# Evaluate each word
for word in words:
# Count each unique word
if word not in newList:
newList.append(word)
answer = newList.sort()
return answer

list.sort sorts the list in place, i.e. it doesn't return a new list. Just write
newList.sort()
return newList

The problem is here:
answer = newList.sort()
sort does not return the sorted list; rather, it sorts the list in place.
Use:
answer = sorted(newList)

Here is an email from Guido van Rossum in Python's dev list explaining why he choose not to return self on operations that affects the object and don't return a new one.
This comes from a coding style (popular in various other languages, I
believe especially Lisp revels in it) where a series of side effects
on a single object can be chained like this:
x.compress().chop(y).sort(z)
which would be the same as
x.compress()
x.chop(y)
x.sort(z)
I find the chaining form a threat to readability; it requires that the
reader must be intimately familiar with each of the methods. The
second form makes it clear that each of these calls acts on the same
object, and so even if you don't know the class and its methods very
well, you can understand that the second and third call are applied to
x (and that all calls are made for their side-effects), and not to
something else.
I'd like to reserve chaining for operations that return new values,
like string processing operations:
y = x.rstrip("\n").split(":").lower()

Python has two kinds of sorts: a sort method (or "member function") and a sort function. The sort method operates on the contents of the object named -- think of it as an action that the object is taking to re-order itself. The sort function is an operation over the data represented by an object and returns a new object with the same contents in a sorted order.
Given a list of integers named l the list itself will be reordered if we call l.sort():
>>> l = [1, 5, 2341, 467, 213, 123]
>>> l.sort()
>>> l
[1, 5, 123, 213, 467, 2341]
This method has no return value. But what if we try to assign the result of l.sort()?
>>> l = [1, 5, 2341, 467, 213, 123]
>>> r = l.sort()
>>> print(r)
None
r now equals actually nothing. This is one of those weird, somewhat annoying details that a programmer is likely to forget about after a period of absence from Python (which is why I am writing this, so I don't forget again).
The function sorted(), on the other hand, will not do anything to the contents of l, but will return a new, sorted list with the same contents as l:
>>> l = [1, 5, 2341, 467, 213, 123]
>>> r = sorted(l)
>>> l
[1, 5, 2341, 467, 213, 123]
>>> r
[1, 5, 123, 213, 467, 2341]
Be aware that the returned value is not a deep copy, so be cautious about side-effecty operations over elements contained within the list as usual:
>>> spam = [8, 2, 4, 7]
>>> eggs = [3, 1, 4, 5]
>>> l = [spam, eggs]
>>> r = sorted(l)
>>> l
[[8, 2, 4, 7], [3, 1, 4, 5]]
>>> r
[[3, 1, 4, 5], [8, 2, 4, 7]]
>>> spam.sort()
>>> eggs.sort()
>>> l
[[2, 4, 7, 8], [1, 3, 4, 5]]
>>> r
[[1, 3, 4, 5], [2, 4, 7, 8]]

Python habitually returns None from functions and methods that mutate the data, such as list.sort, list.append, and random.shuffle, with the idea being that it hints to the fact that it was mutating.
If you want to take an iterable and return a new, sorted list of its items, use the sorted builtin function.

To understand why it does not return the list:
sort() doesn't return any value while the sort() method just sorts the elements of a given list in a specific order - ascending or descending without returning any value.
So problem is with answer = newList.sort() where answer is none.
Instead you can just do return newList.sort().
The syntax of the sort() method is:
list.sort(key=..., reverse=...)
Alternatively, you can also use Python's in-built function sorted() for the same purpose.
sorted(list, key=..., reverse=...)
Note: The simplest difference between sort() and sorted() is: sort() doesn't return any value while, sorted() returns an iterable list.
So in your case answer = sorted(newList).

A small piece of wisdom which I didn't see in other answers:
All methods for mutable objects in python (like lists) which modify the list return None. So, for lists this also includes list.append(), list.reverse(), etc. That's why the syntax should be
myList.sort()
Meanwhile, methods for any immutable object (like strings) must be assigned like so:
myString = myString.strip()

you can use sorted() method if you want it to return the sorted list.
It's more convenient.
l1 = []
n = int(input())
for i in range(n):
user = int(input())
l1.append(user)
sorted(l1,reverse=True)
list.sort() method modifies the list in-place and returns None.
if you still want to use sort you can do this.
l1 = []
n = int(input())
for i in range(n):
user = int(input())
l1.append(user)
l1.sort(reverse=True)
print(l1)