difference between forms of input a list to function

[akbarza]

hi
what is difference between f(b) and f(b[:]) in below? when i run them the outputs are different
1)

b=[1,2]
id(b)
def f(a):
    a[0]=3
    print("a: ",a, "id(a): ",id(a))

f(b)
print("b , id(b)",b,id(b))

2)

b=[1,2]
id(b)
def f(a):
    a[0]=3
    print("a: ",a, "id(a): ",id(a))
f(b[:])
print("b , id(b)",b,id(b))

please explain, thanks

[deanhystad]

b[:] makes a slice of b. The slice has all the elements as b. Did you try looking up "python [:]"?

[snippsat]

The be more accurate when do f(b[:] you are making a copy of b then pass if to f.
func-1 you're passing the reference of the list,so the function modifies the original list.
func-2 you're passing a shallow copy of the list to the function,so the original list remains unchanged.

To give an other example,which also is a common mistake(modify a list you loop over) why is albatross in output?

words = ['eagle', 'albatross', 'mouse', 'dog', 'tiger']
for w in words:
    if len(w) > 3:
        words.remove(w)

print(words)

Output:
['albatross', 'dog']

If loop over copy of words then it work.

words = ['eagle', 'albatross', 'mouse', 'dog', 'tiger']
for w in words[:]:
    if len(w) > 3:
        words.remove(w)

print(words)

Output:
['dog']

Or better do not use copy or remove at all.

words = ['eagle', 'albatross', 'mouse', 'dog', 'tiger']
words = [w for w in words if len(w) <= 3]
print(words)

Output:
['dog']

[deanhystad]

I don't think that is more accurate at all. b.copy(), b[:] and b[0:] all create new lists that are populated with the all the elements from b, so they are all "copies", but the lists returned by b[:] and b[0::] are created through slicing. New python programmers need to learn about list slicing.

[akbarza]

The be more accurate when do f(b[:] you are making a copy of b then pass if to f.
func-1 you're passing the reference of the list,so the function modifies the original list.
func-2 you're passing a shallow copy of the list to the function,so the original list remains unchanged.

To give an other example,which also is a common mistake(modify a list you loop over) why is albatross in output?

words = ['eagle', 'albatross', 'mouse', 'dog', 'tiger']
for w in words:
    if len(w) > 3:
        words.remove(w)

print(words)

Output:
['albatross', 'dog']

If loop over copy of words then it work.

words = ['eagle', 'albatross', 'mouse', 'dog', 'tiger']
for w in words[:]:
    if len(w) > 3:
        words.remove(w)

print(words)

Output:
['dog']

Or better do not use copy or remove at all.

words = ['eagle', 'albatross', 'mouse', 'dog', 'tiger']
words = [w for w in words if len(w) <= 3]
print(words)

Output:
['dog']

hi
I copied and ran the two codes and their outputs was as yours. but I can not understand it.
why in the first code output, there is "albatross " ?
thanks

[snippsat]

why in the first code output, there is "albatross " ?

When modify a list while iterating over it,end up skipping some items in the list(mess up indexing).
First item: eagle' (length > 3) → Remove eagle from the list.
Due to the removal, albatross moves to the first position in the list.
But the loop has already processed the first position, so it moves to the second item, which is now mouse.

In this example are making a new list,and not modifying the original list

words = ['eagle', 'albatross', 'mouse', 'dog', 'tiger']
words = [w for w in words if len(w) <= 3]
print(words)

Output:
['dog']

If write code ove(list comprehension) as a ordinary loop.

words = ['eagle', 'albatross', 'mouse', 'dog', 'tiger']
filtered_words = []
for w in words:
    if len(w) <= 3:
        filtered_words.append(w)

print(filtered_words)

Output:
['dog']

Then is easier to see that we append to new list filtered_words and not modifying the original list words

[bterwijn]

You can use module 'memory_graph':

https://pypi.org/project/memory-graph

to see the call stack. Here stack frame "0: <module>" holds all the global variables and stack frame "1: f" holds the local variables of the f() function.

import memory_graph # see install instructions at link above
id(b)
def f(a):
    a[0]=3
    print("a: ",a, "id(a): ",id(a))
    memory_graph.show( memory_graph.get_call_stack() ) # draw call stack
 
f(b)
print("b , id(b)",b,id(b))

In the first program we see that after calling the f() function variable local variable 'a' shares its data with global variable 'b' of the calling frame. Therefore if 'a' is changed 'b' is also changed.

Now the second program:

import memory_graph
id(b)
def f(a):
    a[0]=3
    print("a: ",a, "id(a): ",id(a))
    memory_graph.show( memory_graph.get_call_stack() ) # draw call stack
    
f(b[:])
print("b , id(b)",b,id(b))

Here f() is called with 'b[:]' (list slicing), so a copy of its data is send to f(). Therefore the local variable 'a' does not share any data with global variable 'b' of the calling frame and when 'a' is changed is does not change 'b'.

Full disclosure: I am the developer of memory_graph.