First Python list index greater than x?

Question

What would be the most Pythonic way to find the first index in a list that is greater than x?

For example, with

list = [0.5, 0.3, 0.9, 0.8]


        

Answer 1

for index, elem in enumerate(elements):
    if elem > reference:
        return index
raise ValueError("Nothing Found")

Answer 2

Try this one:

def Renumerate(l):
    return [(len(l) - x, y) for x,y in enumerate(l)]

example code:

Renumerate(range(10))

output:

(10, 0)
(9, 1)
(8, 2)
(7, 3)
(6, 4)
(5, 5)
(4, 6)
(3, 7)
(2, 8)
(1, 9)

Answer 3

>>> alist= [0.5, 0.3, 0.9, 0.8]
>>> [ n for n,i in enumerate(alist) if i>0.7 ][0]
2

Answer 4

You could also do this using numpy:

import numpy as np

list(np.array(SearchList) > x).index(True)

Answer 5

1) NUMPY SOLUTION, general lists

If you are happy to use numpy, then the following will work on general lists (sorted or unsorted):

numpy.argwhere(np.array(searchlist)>x)[0]

or if you need the answer as a list:

numpy.argwhere(np.array(searchlist)>x).tolist()[0]

or if you need the answer as a integer index:

numpy.argwhere(np.array(searchlist)>x).tolist()[0][0]

2) NUMPY SOLUTION, sorted lists

However, if your search list is sorted, it is much cleaner and nicer to use the function np.searchsorted:

numpy.searchsorted(searchlist,x)

The nice thing about using this function is that as well as search for a single value x, you can also return a list of indices for a list of values x, (and it is very efficient relative to a list comprehension in this case).

Answer 6

Another one:

map(lambda x: x>.7, seq).index(True)