Skip over a value in the range function in python

Question

What is the pythonic way of looping through a range of numbers and skipping over one value? For example, the range is from 0 to 100 and I would like to skip 50.

Edit: Here's the code that I'm using

for i in range(0, len(list)):
    x= listRow(list, i)
    for j in range (#0 to len(list) not including x#)
        ...

I could do that but is there any way to get it in the structure of the loop itself? — David, Jun 06 '14 at 20:15

score 107 · Accepted Answer · edited May 10 '17 at 09:31

107

You can use any of these:

# Create a range that does not contain 50
for i in [x for x in xrange(100) if x != 50]:
    print i

# Create 2 ranges [0,49] and [51, 100] (Python 2)
for i in range(50) + range(51, 100):
    print i

# Create a iterator and skip 50
xr = iter(xrange(100))
for i in xr:
    print i
    if i == 49:
        next(xr)

# Simply continue in the loop if the number is 50
for i in range(100):
    if i == 50:
        continue
    print i

edited May 10 '17 at 09:31

Right leg

16,080
7
48
81

answered Jun 06 '14 at 20:18

njzk2

38,969
7
69
107

The third suggestion throws TypeError because you did not explicitly create an iterator – aestrivex Jun 06 '14 at 20:22
(1) creates two lists, (2) concatenates two lists, (3) does not work, xrange is not a iterator. (4) should use xrange to avoid creating a list, and is the so far best solution. – Daniel Jun 06 '14 at 20:23
2

How does (1) create two lists? `xrange(100)` is not a list. And you can avoid creating the second list by returning a generator instead: `for i in (x for x in xrange(100) if x is not 50)` – aestrivex Jun 06 '14 at 20:25
@Daniel: 1/ creates one list only because of xrange. 2/ I know, not the most efficient, but clear. 3/ typo, thanks for pointing out 4/ probably, yes. (but for a 100 elements it is fine) – njzk2 Jun 06 '14 at 20:26
4

#2 with Python 3.x: `...list(range(50)) + list(range(51, 100)):` – loxosceles Feb 15 '17 at 22:17
1

@loxosceles or using `itertools.chain(range(50), range(51, 100))` instead, to avoid creating lists (although my initial solution is indeed creating 2 lists) – njzk2 Feb 15 '17 at 23:09
#2 doesn't work in Python 3. The rest are all quite inefficient. – Asclepius Sep 10 '19 at 14:17
1

@Acumenus this question is tagged `python`, not `python3`. If you have more efficient solutions, please share them so everyone can benefit :) – njzk2 Sep 12 '19 at 01:54

score 12 · Answer 2 · answered Oct 06 '18 at 15:09

In addition to the Python 2 approach here are the equivalents for Python 3:

# Create a range that does not contain 50
for i in [x for x in range(100) if x != 50]:
    print(i)

# Create 2 ranges [0,49] and [51, 100]
from itertools import chain
concatenated = chain(range(50), range(51, 100))
for i in concatenated:
    print(i)

# Create a iterator and skip 50
xr = iter(range(100))
for i in xr:
    print(i)
    if i == 49:
        next(xr)

# Simply continue in the loop if the number is 50
for i in range(100):
    if i == 50:
        continue
    print(i)

Ranges are lists in Python 2 and iterators in Python 3.

Asclepius · Answer 3 · 2023-02-26T03:36:50.567

2

If the index of the skipped value is known, it is not necessary to compare each number:

import itertools

m, n = 5, 10
for i in itertools.chain(range(m), range(m + 1, n)):
    print(i)  # skips m = 5

As an aside, you woudn't want to use (*range(m), *range(m + 1, n)) even though it works because it will expand the iterables into a tuple and this is memory inefficient.

Credit: comment by njzk2

edited Feb 26 '23 at 03:36

answered Sep 10 '19 at 14:20

Asclepius

57,944
17
167
143

score 1 · Answer 4 · answered Oct 04 '19 at 07:52

1

for i in range(0, 101):
if i != 50:
    do sth
else:
    pass

answered Oct 04 '19 at 07:52

Maria

11
1

score 0 · Answer 5 · answered Jun 06 '14 at 20:13

0

for i in range(100):
    if i == 50:
        continue
    dosomething

answered Jun 06 '14 at 20:13

Daniel

42,087
4
55
81

score 0 · Answer 6 · answered Jun 06 '14 at 20:14

It depends on what you want to do. For example you could stick in some conditionals like this in your comprehensions:

# get the squares of each number from 1 to 9, excluding 2
myList = [i**2 for i in range(10) if i != 2]
print(myList)

# --> [0, 1, 9, 16, 25, 36, 49, 64, 81]

jawad846 · Answer 7 · 2021-04-19T13:16:57.183

0

This works for me;

example:

x = ['apple', 'orange', 'grape', 'lion', 'banana', 'watermelon', 'onion', 'cat',]

for xr in x:
    if xr in 'onion':
        print('onion is a vegetable')
        continue
    if (xr not in 'lion' and xr not in 'cat'):
        print(xr, 'is a fruit')

Output -->

apple is a fruit
orange is a fruit
grape is a fruit
banana is a fruit
watermelon is a fruit
onion is a vegetable

edited Apr 19 '21 at 13:16

answered Apr 19 '21 at 10:52

jawad846

683
1
9
21

score 0 · Answer 8 · answered Mar 02 '23 at 10:39

If you would like to skip over consecutive numbers where the initial number may be unknown, it is best to use a while loop.

# Loop through a range of numbers from 0 to 100 and skip over number 50 as well
# as the next 3 consecutive numbers (51, 52, 53). Additionally, skip over 11 to 14.

i = 0
initial_skip_num = (50, 11, )
while i < 100:
    if i in initial_skip_num:
        i += 4
    print(i)
    i += 1

Output:

0
1
...
9
10
15
16
...
48
49
54
55
...

This is helpful if the initial number to skip is unknown or multiple consecutive numbers need to be skipped

score -1 · Answer 9 · answered Nov 06 '16 at 16:11

what you could do, is put an if statement around everything inside the loop that you want kept away from the 50. e.g.

for i in range(0, len(list)):
    if i != 50:
        x= listRow(list, i)
        for j in range (#0 to len(list) not including x#)

Skip over a value in the range function in python

9 Answers9

Linked