Because it's more common to call range(0, 10) which returns [0,1,2,3,4,5,6,7,8,9] which contains 10 elements which equals len(range(0, 10)). There's a tendency in programming to use 0-based indexing.
Also, consider the following common code snippet:
for i in range(len(li)):
pass
Could you see that if range() went up to exactly len(li) that this would be problematic? The programmer would need to explicitly subtract 1. This also follows the common trend of programmers preferring for(int i = 0; i < 10; i++) over for(int i = 0; i <= 9; i++).
If you are calling range with a start of 1 frequently, you might want to define your own function:
>>> def range1(start, end):
... return range(start, end+1)
...
>>> range1(1, 10)
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Because it's more common to call range(0, 10) which returns [0,1,2,3,4,5,6,7,8,9] which contains 10 elements which equals len(range(0, 10)). There's a tendency in programming to use 0-based indexing.
Also, consider the following common code snippet:
for i in range(len(li)):
pass
Could you see that if range() went up to exactly len(li) that this would be problematic? The programmer would need to explicitly subtract 1. This also follows the common trend of programmers preferring for(int i = 0; i < 10; i++) over for(int i = 0; i <= 9; i++).
If you are calling range with a start of 1 frequently, you might want to define your own function:
>>> def range1(start, end):
... return range(start, end+1)
...
>>> range1(1, 10)
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Although there are some useful algorithmic explanations here, I think it may help to add some simple 'real life' reasoning as to why it works this way, which I have found useful when introducing the subject to young newcomers:
With something like 'range(1,10)' confusion can arise from thinking that pair of parameters represents the "start and end".
It is actually start and "stop".
Now, if it were the "end" value then, yes, you might expect that number would be included as the final entry in the sequence. But it is not the "end".
Others mistakenly call that parameter "count" because if you only ever use 'range(n)' then it does, of course, iterate 'n' times. This logic breaks down when you add the start parameter.
So the key point is to remember its name: "stop". That means it is the point at which, when reached, iteration will stop immediately. Not after that point.
So, while "start" does indeed represent the first value to be included, on reaching the "stop" value it 'breaks' rather than continuing to process 'that one as well' before stopping.
One analogy that I have used in explaining this to kids is that, ironically, it is better behaved than kids! It doesn't stop after it supposed to - it stops immediately without finishing what it was doing. (They get this ;) )
Another analogy - when you drive a car you don't pass a stop/yield/'give way' sign and end up with it sitting somewhere next to, or behind, your car. Technically you still haven't reached it when you do stop. It is not included in the 'things you passed on your journey'.
I hope some of that helps in explaining to Pythonitos/Pythonitas!
Why Range() does not include last or end element ? Is it possible to include last element of range?
Including the last value in a range loop Python 3 - Stack Overflow
python - For loop range and interval, how to include last step - Stack Overflow
list - How do I include the last element in a range in python? - Stack Overflow
I think you just need to make sure you are adding 1 to the ordinal, not the char. Additionally you don't need the inner for loop, all you need to do is cast the current index back to a character with chr():
for char in range(ord(start_character), ord(end_character)+1):
print(chr(char))
In the range loop you can add the +1 to the end_character like so:
for char in range(ord(start_character), ord(end_character) + 1):
ord in python turns the character into an integer.
This is impossible to do with Python's range. But this can be accomplished by creating your own generator function.
def myRange(start,end,step):
i = start
while i < end:
yield i
i += step
yield end
for i in myRange(0,99,20):
print(i)
Output:
0
20
40
60
80
99
First of all, it usually does not makes much sense to include the end condition, the idea of a range is to perform hops until the end value is reached.
Nevertheless, you can for instance use itertools.chain for that:
from itertools import chain
for i in chain(range(0,99,20), [99]):
# ...
passchain concatenates iterables together. So after the range(..) is exhausted, it will iterate over the next iterable, and so on.
The above approach is not very elegant: it requires some knowledge about how chain works. We can however encapsulate that logic:
def range_with_end(start, stop, step):
return chain(range(start, stop, step), (stop,))these are three solutions for your usage
x=[3,4,5,6]
total=0
#solution 1 :
total =sum(x)
#solution 2
for i in x:
total= total + i
print(total)
#solution 3
for i in range(0,len(x)):
total= total + x[i]
print(total)
In your code the dereferenced for-loop looks like this:
for i in range(3,6):
So it loops through with i values of 3, 4 and 5.
To fix it you could either add +1 to your range if the numbers in your list are always ascending by 1, its not good code though in my opinion.
for i in range (x[0],x[-1]+1):
I'd recommend you to not dereference the list in a range function in the first place, a better option would be the use of len():
for i in range(len(x)):
total = total + x[i]
That way you're making sure to loop through the whole list and reference every element.
Note that this was more for explanatory reasons, the cleanest way is as stated in other answers:
for i in x:
total += i
Range returns a list counting from 0 to 2 (3 minus 1) ([0, 1, 2]), so you can simply access the element directly from the range element. Index -1 refers to the last element.
print(range(3)[-1])
I the range is in your collection you should be able to reference that object directly by the index.
if len(j) > 2:
print("Value: " + j[2])
well, from the help:
>>> help(range)
range(...)
range([start,] stop[, step]) -> list of integers
Return a list containing an arithmetic progression of integers.
range(i, j) returns [i, i+1, i+2, ..., j-1]; start (!) defaults to 0.
When step is given, it specifies the increment (or decrement).
For example, range(4) returns [0, 1, 2, 3]. The end point is omitted!
These are exactly the valid indices for a list of 4 elements.
so the last increment is not stop, but the last step before stop.
- in countMe shouldn't the code go up till 18 ;
- why is the last number printed in countMe 15, and not 18 ;
- why is that in the second function oddsOut the function only founts till 7 for j and not 8 even though j is 8 ;
- why is the last number printed in oddsOut 14.
more generally speaking the answer to those questions is that in most of the languages, a range is defined as [start:stop[, i.e. the last value of the range is never included, and the indexes start always at 0. The mess being that in a few languages and when working on algorithmics, ranges start at 1 and are inclusive with the last value.
In the end, if you want to include the last value you can do:
def closed_range(start, stop, step=1):
return range(start, stop+1, step)
or in your example:
>>> def countMe(num):
>>> for i in range(0, num+1, 3):
>>> print (i)
>>>
>>> countMe(18)
0
3
6
9
12
15
18
>>>
The stop parameter in a range does not include that number for example
for i in range(0,5):
print i
would print 0-4 but not 5.
I first thought that it just returns numbers and you can do a lot of stuff with those numbers but now I am seeing that it is being used to run a loop a specific number of times. It's just not making sense to me why this works, like I can't intuitively understand it.
Tho I have started using in my code, an explanation on why it was developed this way would help.
edit: forgot to mention I want the indeces not just the items
the methods I can think of are:
for i in range(len(arr) - 1)
for i, e in enumerate(arr[:len(arr) - 1])
I know range(len(arr)) is frowned upon, but I don't see how it's worse than enumerate in this case. In fact using enumerate on a shortened list and calling len to find the second to last element of that list seems extremely clunky and far less readable.
What's the best practice for doing this?