Detect If Item is the Last in a List [duplicate]

Answer 1

Test if the index value is the length of the list - 1

Like this:

    if item != (len(a) - 1):
        #whatever you wanted to do

    #or, if you want the index to be the last...
    if item == (len(a) - 1):
        #whatever you wanted to do

Answer 2

You are not getting the expected output because in your code you say, "Hey python, print a comma if current element is not equal to last element."

Second element 9 is equal to last element, hence the comma is not printed.

The right way to check for the last element is to check the index of element :

a = ['hello', 9, 3.14, 9]
for item in a:
    print(item, end='')
    if a.index(item) != len(a)-1: #<--------- Checking for last element here
        print(', ')

(Also, this might throw a value error. You should check for that too.)

Answer 3

If you really just want to join your elements into a comma+newline-separated string then it's easier to use str.join method:

elements = ['hello', 9, 3.14, 9]
s = ',\n'.join(str(el) for el in elements)
# And you can do whatever you want with `s` string:
print(s)

Answer 4

If you are just looking to see if the item is last in the (or any item in any list) you could try using a function to check if an item is last in a list.

a = ['hello', 9, 3.14, 9]
def is_last(alist,choice):
        if choice == alist[-1]:
            print("Item is last in list!")
            return True
        else:
            print("Item is not last")
            return False

if is_last(a,9) == True:
    <do things>

else:
    <do other things>

Answer 5

While these answers may work for the specific case of the OP, I found they were unsatisfactory for broader application.

Here are the methods I could think of/saw here and their respective timings.

Index Method

urlist_len = len(urlist)-1
for x in urlist:
    if urlist.index(x) == urlist_len:
        pass

Negative Slice Method

for x in urlist:
    if x == urlist[-1]:
        pass

Enumerate Method

urlist_len = len(urlist)-1
for index, x in enumerate(urlist):
    if index == urlist_len:
        pass

Here are some timing for some different methods:

╔════════════════════════════════════════════════════════════════╗
║                       Timing Results (s)                       ║
╠═══════════════════════╦════╦═════════╦════════╦═══════╦════════╣
║ List size             ║ 20 ║ 200     ║ 2000   ║ 20000 ║ 200000 ║
╠═══════════════════════╬════╬═════════╬════════╬═══════╬════════╣
║                       ║ 0  ║ 0.0006  ║ 0.051  ║ 5.2   ║ 560    ║
║ Index Method          ║    ║         ║        ║       ║        ║
╠═══════════════════════╬════╬═════════╬════════╬═══════╬════════╣
║                       ║ 0  ║ 0       ║ 0.0002 ║ 0.003 ║ 0.034  ║
║ Negative Slice Method ║    ║         ║        ║       ║        ║
╠═══════════════════════╬════╬═════════╬════════╬═══════╬════════╣
║ Enumerate Method      ║ 0  ║ 0.00004 ║ 0.0005 ║ 0.016 ║ 0.137  ║
╚═══════════════════════╩════╩═════════╩════════╩═══════╩════════╝

^{Note: values <10us rounded to 0}

As you can see, the index method is always slower, and it only get exponentially worse as the list size increases. I don't see any reason to use it ever. The Negative slice method is the fastest in all cases, but if you have duplicate items in your list, it will give you a false positive. Also, the negative slice method requires that the sequence you are iterating over supports indexing. So, in the case of duplicate items in your list (or not index supporting sequence) use the fast-but-not-fastest enumerate method.

Edit: as a commentator noted, calculating the length of the list within the loop isn't ideal. I was able to shave 35% off the enumerate method (not reflected in the table at this moment) using this knowledge.

---

tldr: use negative slice if all elements are unique and sequence supports indexing, otherwise enumerate method

Answer 6

Rather than try and detect if you are at the last item, print the comma and newline when printing the next (which only requires detecting if you are at the first):

a = ['hello', 9, 3.14, 9]
for i, item in enumerate(a):
    if i:  # print a separator if this isn't the first element
        print(',')
    print(item, end='')
print()  # last newline

The enumerate() function adds a counter to each element (see What does enumerate mean?), and if i: is true for all values of the counter except 0 (the first element).

Or use print() to insert separators:

print(*a, sep=',\n')

The sep value is inserted between each argument (*a applies all values in a as separate arguments, see What does ** (double star) and * (star) do for parameters?). This is more efficient than using print(',n'.join(map(str, a))) as this doesn't need to build a whole new string object first.