Python "R Cut" function

Question

I am wondering if there is a function in Python already written for the goal that I describe later. If not, what would be the easiest way to implement. My code is attached.

Say I have a range from 1 to 999999999. Given a list of numbers like this:

[9, 44, 99]

It would return

[(1,9), (10,44), (45,99), (100, 999999999)]

If the number which are the limits are included in the input numbers, it should handle that also. Say input is

[1, 9, 44, 999999999]

The return should be:

[(1,9), (10,44), (45, 999999999)]

I could write a for loop comparing with a few conditional statement but wondering if there is a more 'smart way'.

Some data massage that might be helpful:

points = [1, 9, 44, 99]
points = sorted(list(set(points + [1, 999999999])))

UPDATED INFO: FINAL CREDITS GIVEN TO alecxe, thanks for your inspiring list comprehension solution

l = sorted(list(set(points + [1, 999999999])))
[(l[i] + int(i != 0), l[i + 1]) for i in xrange(len(l) - 1)]

You can put all that in one line but I think that is unnessary.

Answer 1

pandas.cut()

Example

[1,2,3,4,5,6,7,8,9,10] ---> [A,A,B,B,C,C,D,D,E,E]

R:

x  <- seq(1,10,1)
cut(x, breaks = seq(0,10,2), labels = c('A','B','C','D','E'))

Python:

import pandas
x = range(1, 11, 1)
pandas.cut(x, bins=range(0, 12, 2), labels=['A','B','C','D','E'])