# Skip over a value in the range function in python

It is time inefficient to compare each number, needlessly leading to a linear complexity. Having said that, this approach avoids any inequality checks:

```
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, answer by Locke

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.

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

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
```