+ 1
What is going wrong ??
def lastOccr(list,n): last = list[0] for i in list: if i == n: last = i return list.index(last) list = [0,1,5,9,5,6,5,0,4,1] print(lastOccr(list,5))
8 Answers
+ 4
You can use enumerate() to get any position of the same element in a list.
def lastOccr(list,n):
last = list[0]
for i,e in enumerate(list):
if e == n:
last = i
return last
list = [0,1,5,9,5,6,5,0,4,1]
print(lastOccr(list,5))
+ 3
Adi
5 == 5 so last = 5
And list.index(5) = 2
#this is the index of first 5 not last 5
+ 3
Adi
You can get all indexes and last index using this logic.
#What is going wrong ??
def lastOccr(list, n):
return [i for i, x in enumerate(list) if x == n]
list = [0,1,5,9,5,6,5,0,4,1]
indexes = lastOccr(list, 5)
print (indexes) #all indexes
print (indexes[-1]) #last index
+ 2
3 versions as the comments show:
# version using list iteration
def lastOccr1(lst_, number):
for ndx, num in enumerate(lst_):
if num == number:
ndx_pos = ndx
return ndx_pos
# version using string with rindex() method, thst starts searching from the right side of string
def lastOccr2(lst_, number):
return ''.join(map(str, lst_)).rindex(str(number))
# using enumerate to create the index values, then use max() function
def lastOccr3(lst_, number):
return max([ndx for ndx, num in enumerate(lst_) if num == number])
lst = [0,1,5,9,5,6,5,0,4,1]
print(lastOccr1(lst, 5))
print(lastOccr2(lst, 5))
print(lastOccr3(lst, 5))
+ 1
What is your expected output..
list.index() returns first occurence index
+ 1
So how can i get the last occr index?
+ 1
Try to pass your list but reversed.
lastOccr(list[::-1], 5)
In a reverse list, the index of the first occurrence of n will be the index of the last in your standard list.
However, this method requires an extra step in which you must calculate the real index with the reversed index and your original list's length.
+ 1
#your way modified
def lastOccr(list,n):
last = -1
for i in range(len(list)):
if list[i] == n:
last = i
return last
list = [0,1,5,9,5,6,5,0,4,1]
print(lastOccr(list,5))
#Or
print(len(list) - list[::-1].index(5) -1)
#this works fine if element is in list, else throws error... hope it helps..