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Python list index from a certain point onwards

For a list
a= [1,2,5,3,7,2,3,1,7,4,2,3,4,2,1]
i know that if i do this
a.index(2)
1
However I want want to know what is the cleanest way to find a number from a certain point.
a.index(2,from point 2)
5

The list.index [doc] function has extra parameters:
list.index(x[, start[, end]])
Return zero-based index in the list of the first item whose value is x. Raises a ValueError if there is no such item.
The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list. The returned index is computed relative to the beginning of the full sequence rather than the start argument.
So you can use:
a.index(2, 2)
to start searching from index 2 (including index 2).

via implementation of list.index, you can pass your "point" to achieve your desired output:
a= [1,2,5,3,7,2,3,1,7,4,2,3,4,2,1]
print(a.index(2, 2))
Output:
5
However, another possible solution is to built a dictionary:
val = 2
a= [1,2,5,3,7,2,3,1,7,4,2,3,4,2,1]
locations = dict(enumerate([i for i, c in enumerate(a) if c == val], start=1))
print(locations.get(2, False))
Output:
5

What is fastest way to determine numbers are within specific range of each other in Python?

I have list of numbers as follows -
L = [ 1430185458, 1430185456, 1430185245, 1430185246, 1430185001 ]
I am trying to determine which numbers are within range of "2" from each other. List will be in unsorted when I receive it.
If there are numbers within range of 2 from each other I have to return "1" at exact same position number was received in.
I was able to achieve desired result , however code is running very slow. My approach involves sorting list, iterating it twice taking two pointers and comparing it successively. I will have millions of records coming as seperate lists.
Just trying to see what is best possible approach to address this problem.
Edit - Apology as I was away for a while. List can have any number of elements in it ranging from 1 to n. Idea is to return either 0 or 1 in exact same position number was received. I can not post actual code I implemented but here is pseudo code.
a. create new list as list of list with second part as 0 for each element. We assume that there are no numbers within range of 2 of each other.
[[1430185458,0], [1430185456,0], [1430185245,0], [1430185246,0], [1430185001,0]]
b. sort original list
c. compare first element to second, second to third and so on until end of list is reached and whenever difference is less than or equal to 2 update corresponding second elements in step a to 1.
[[1430185458,1], [1430185456,1], [1430185245,1], [1430185246,1], [1430185001,0]]

The goal is to be fast, so that presumably means an O(N) algorithm. Building an NxN difference matrix is O(N^2), so that's not good at all. Sorting is O(N*log(N)), so that's out, too. Assuming average case O(1) behavior for dictionary insert and lookup, the following is an O(N) algorithm. It rips through a list of a million random integers in a couple of seconds.
def in_range (numbers) :
result = [0] * len(numbers)
index = {}
for idx, number in enumerate(numbers) :
for offset in range(-2,3) :
match_idx = index.get(number+offset)
if match_idx is not None :
result[match_idx] = result[idx] = 1
index[number] = idx
return result
Update
I have to return "1" at exact same position number was received in.
The update to the question asks for a list of the form [[1,1],[2,1],[5,0]] given an input of [1,2,5]. I didn't do that. Instead, my code returns [1,1,0] given [1,2,5]. It's about 15% faster to produce that simple 0/1 list compared to the [[value,in_range],...] list. The desired list can easily be created using zip:
zip(numbers,in_range(numbers)) # Generator
list(zip(numbers,in_range(numbers))) # List of (value,in_range) tuples

I think this does what you need (process() modifies the list L). Very likely it's still optimizable, though:
def process(L):
s = [(v,k) for k,v in enumerate(L)]
s.sort()
j = 0
for i,v_k in enumerate(s):
v = v_k[0]
while j < i and v-s[j][0]>2:
j += 1
while j < i:
L[s[j][1]] = 1
L[s[i][1]] = 1
j += 1

Adding Numbers to a list using variable

I'm doing a programming question and I need a bit of help.
I have a variable where I type in a number and then that number and all the numbers before it go in a list.
For example:
I pick 10 and put it in a variable
And all numbers from 1 - 10 are put in a list
Here is my code:
Top_Num = int(input('Top Number'))
Nums = []
Now lets say I chose 10 for the Top_Num, how do I put 10 numbers into the list?
Thanks.

You can actually use Pythons built in range(int) function to do exactly this.
If you want the array to start at 1 and include the input number, you can use
Nums = list(range(1, Top_Num + 1))
The first argument of 1 indicates the start value of the array, and the second argument Top_Num + 1 is the number that the array goes up to (exclusive).

Nums = [num for num in range(1, Top_Num + 1)]
It uses list comprehensions too, which is (kinda) an important concept in python.

How to change for-loop iterator variable in the loop in Python?

I want to know if is it possible to change the value of the iterator in its for-loop?
For example I want to write a program to calculate prime factor of a number in the below way :
def primeFactors(number):
for i in range(2,number+1):
if (number%i==0)
print(i,end=',')
number=number/i
i=i-1 #to check that factor again!
My question : Is it possible to change the last two line in a way that when I change i and number in the if block, their value change in the for loop!
Update: Defining the iterator as a global variable, could help me? Why?

Short answer (like Daniel Roseman's): No
Long answer: No, but this does what you want:
def redo_range(start, end):
while start < end:
start += 1
redo = (yield start)
if redo:
start -= 2
redone_5 = False
r = redo_range(2, 10)
for i in r:
print(i)
if i == 5 and not redone_5:
r.send(True)
redone_5 = True
Output:
3
4
5
5
6
7
8
9
10
As you can see, 5 gets repeated. It used a generator function which allows the last value of the index variable to be repeated. There are simpler methods (while loops, list of values to check, etc.) but this one matches you code the closest.

No.
Python's for loop is like other languages' foreach loops. Your i variable is not a counter, it is the value of each element in a list, in this case the list of numbers between 2 and number+1. Even if you changed the value, that would not change what was the next element in that list.

The standard way of dealing with this is to completely exhaust the divisions by i in the body of the for loop itself:
def primeFactors(number):
for i in range(2,number+1):
while number % i == 0:
print(i, end=',')
number /= i
It's slightly more efficient to do the division and remainder in one step:
def primeFactors(number):
for i in range(2, number+1):
while True:
q, r = divmod(number, i)
if r != 0:
break
print(i, end=',')
number = q

The only way to change the next value yielded is to somehow tell the iterable what the next value to yield should be. With a lot of standard iterables, this isn't possible. however, you can do it with a specially coded generator:
def crazy_iter(iterable):
iterable = iter(iterable)
for item in iterable:
sent = yield item
if sent is not None:
yield None # Return value of `iterable.send(...)`
yield sent
num = 10
iterable = crazy_iter(range(2, 11))
for i in iterable:
if not num%i:
print i
num /= i
if i > 2:
iterable.send(i-1)
I would definitely not argue that this is easier to read than the equivalent while loop, but it does demonstrate sending stuff to a generator which may gain your team points at your next local programming trivia night.

It is not possible the way you are doing it. The for loop variable can be changed inside each loop iteration, like this:
for a in range (1, 6):
print a
a = a + 1
print a
print
The resulting output is:
1
2
2
3
3
4
4
5
5
6
It does get modified inside each for loop iteration.
The reason for the behavior displayed by Python's for loop is that, at the beginning of each iteration, the for loop variable is assinged the next unused value from the specified iterator. Therefore, whatever changes you make to the for loop variable get effectively destroyed at the beginning of each iteration.
To achieve what I think you may be needing, you should probably use a while loop, providing your own counter variable, your own increment code and any special case modifications for it you may need inside your loop. Example:
a = 1
while a <= 5:
print a
if a == 3:
a = a + 1
a = a + 1
print a
print
The resulting output is:
1
2
2
3
3
5
5
6

Yes, we can only if we dont change the reference of the object that we are using. If we can edit the number by accessing the reference of number variable, then what you asked is possible.
A simple example:
a=[1,2,3]
a=a+[4]==>here, a new object is created which plots to different address.
a+=[4]==>here , the same object is getting updated which give us the desired result.
number=10
list1=list(range(2,number+1))
# list1
for i in list1:
print(list1,i)
if (number%i==0):
print(i,end=',')
number=number//i #we can simply replace it with number//=i to edit the number without changing the reference or without creating a new object.
try:
[list1.pop() for i in range(10,0,-1) if(i>number)]
#here pop() method is working on the same object which list created by number refers. so, we can able to change the iterable in the forloop.
except:
continue
i=i-1 #to check that factor again!

Index error:list assignment index out of range

I want to:
Take two inputs as integers separated by a space (but in string form).
Club them using A + B.
Convert this A + B to integer using int().
Store this integer value in list C.
My code:
C = list()
for a in range(0, 4):
A, B = input().split()
C[a] = int(A + B)
but it shows:
IndexError: list assignment index out of range
I am unable understand this problem. How is a is going out of the range (it must be starting from 0 ending at 3)?
Why it is showing this error?

Why your error is occurring:
You can only reference an index of a list if it already exists. On the last line of every iteration you are referring to an index that is yet to be created, so for example on the first iteration you are trying to change the index 0, which does not exist as the list is empty at that time. The same occurs for every iteration.
The correct way to add an item to a list is this:
C.append(int(A + B))
Or you could solve a hella lot of lines with an ultra-pythonic list comprehension. This is built on the fact you added to the list in a loop, but this simplifies it as you do not need to assign things explicitly:
C = [sum(int(item) for item in input("Enter two space-separated numbers: ").split()) for i in range(4)]
The above would go in place of all of the code that you posted in your question.

The correct way would be to append the element to your list like this:
C.append(int(A+B))
And don't worry about the indices

Here's a far more pythonic way of writing your code:
c = []
for _ in range(4): # defaults to starting at 0
c.append(sum(int(i) for i in input("Enter two space-separated numbers").split()))
Or a nice little one-liner:
c = [sum(int(i) for i in input("Enter two space-separated numbers").split()) for _ in range(4)]