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#)
        ...

Asked By: David

||

Source

Answer 1

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

Answered By: Daniel

Answer 2

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]

Answered By: MxLDevs

Answer 3

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

Answered By: njzk2

Answer 4

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#)

Answered By: Anonemuss

Answer 5

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.

Answered By: Smittie

Answer 6

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

Answered By: Asclepius

Answer 7

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

Answered By: Maria

Answer 8

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

Answered By: Mohamed Jawad

Skip over a value in the range function in python

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