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how can a perfect fourth interval be considered either consonant or dissonant?

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Define a list range inside a list

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box;
}

9

in python is there a way to create of list that will skip numbers and will continue after skipping? something like the following code:

x = [1...3, 6...10]

print(x)

# [1,2,3,6,7,8,9,10]

Well its easy to write a for loop and then skip each defined index/value, or i can just use range, what I am looking for is a shorter more readable line. If not I can understand.

python

share|improve this question

edited 9 hours ago

Led

asked 9 hours ago

LedLed

424515

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Possible duplicate of How can I generate a list of consecutive numbers?
  
  – ivan_pozdeev
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  No @ivan_pozdeev, the suggested dupe does not consider multiple slice objects, rather a simple continuous range
  
  – yatu
  9 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Python doesn't support any special range syntax, never mind implicit concatenation of such within the same list literal.
  
  – chepner
  2 hours ago
  

  
  
  
  

add a comment | 

9

in python is there a way to create of list that will skip numbers and will continue after skipping? something like the following code:

x = [1...3, 6...10]

print(x)

# [1,2,3,6,7,8,9,10]

Well its easy to write a for loop and then skip each defined index/value, or i can just use range, what I am looking for is a shorter more readable line. If not I can understand.

python

share|improve this question

edited 9 hours ago

Led

asked 9 hours ago

LedLed

424515

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Possible duplicate of How can I generate a list of consecutive numbers?
  
  – ivan_pozdeev
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  No @ivan_pozdeev, the suggested dupe does not consider multiple slice objects, rather a simple continuous range
  
  – yatu
  9 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Python doesn't support any special range syntax, never mind implicit concatenation of such within the same list literal.
  
  – chepner
  2 hours ago
  

  
  
  
  

add a comment | 

in python is there a way to create of list that will skip numbers and will continue after skipping? something like the following code:

x = [1...3, 6...10]

print(x)

# [1,2,3,6,7,8,9,10]

Well its easy to write a for loop and then skip each defined index/value, or i can just use range, what I am looking for is a shorter more readable line. If not I can understand.

python

share|improve this question

edited 9 hours ago

Led

asked 9 hours ago

LedLed

424515

x = [1...3, 6...10]

print(x)

# [1,2,3,6,7,8,9,10]

share|improve this question

edited 9 hours ago

Led

asked 9 hours ago

LedLed

424515

share|improve this question

edited 9 hours ago

Led

asked 9 hours ago

LedLed

424515

asked 9 hours ago

LedLed

424515

asked 9 hours ago

LedLed

424515

4 Answers
4

active

oldest

votes

13

Simplest way to do this is to call range() and unpack result inside list assignment.

x = [*range(1, 4), *range(6, 11)]

share|improve this answer

edited 9 hours ago

answered 9 hours ago

Olvin RoghtOlvin Roght

34219

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Please name and refer to the docs of the corresponding syntax constructs
  
  – ivan_pozdeev
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  whats with the *?
  
  – Led
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Exactly this is the way to go. gj
  
  – Netwave
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @Led, Unpacking arguments
  
  – Olvin Roght
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  While this works, it's worth noting that this creates the entire list in memory. If you just want to iterate over the resulting sequence once, then this will consume more memory than the itertools.chain method. 40 bytes per number on my system, to be exact. This is fine for small sequences, but if your sequence has a billion numbers, well, you do the math...
  
  – marcelm
  6 hours ago
  

  
  
  
  

 | 
show 1 more comment

11

Alternatively you can use itertools.chain:

>>> import itertools

>>> list(itertools.chain(range(1, 5), range(20, 25)))

[1, 2, 3, 4, 20, 21, 22, 23, 24]

share|improve this answer

answered 9 hours ago

NetwaveNetwave

14.1k22247

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  interesting, but is there no native way? i mean I need to import something? thank you very much
  
  – Led
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  @Led, itertools is one of the most powerful libs that come with python, it is as native as it can be. Otherwise the unpacking sintax is handy also.
  
  – Netwave
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @Led why does it matter if you have to import from the standard library? It's part of Python
  
  – juanpa.arrivillaga
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  im looking for something short without importing.
  
  – Led
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Led Importing is your friend. Learn that sooner rather than later.
  
  – wizzwizz4
  6 hours ago
  

  
  
  
  

add a comment | 

5

If numpy is an option, you can use np.r_ to concatenate slice objects:

import numpy as np

np.r_[range(1,4), range(6,11)]

# array([ 1,  2,  3,  6,  7,  8,  9, 10])

share|improve this answer

answered 9 hours ago

yatuyatu

15.8k41642

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  And since we just need slice objects you can write np.r_[1:4, 6:11] as well.
  
  – miradulo
  2 hours ago
  

  
  
  
  

add a comment | 

2

You can turn it into a recursive function:

def recursive_ranges(ranges):

    if len(ranges) == 1:

        return list(range(*ranges[0]))

    else:

        return list(range(*ranges[0])) + recursive_ranges(ranges[1:])

You can then call this, specifying ranges as a list of lists:

ranges = [[1, 4], [6, 11]]

recursive_ranges(ranges)

# [1, 2, 3, 6, 7, 8, 9, 10]

Note the *ranges[0] is used to unpack the elements in ranges[0] into individual arguments. Essentially the recursive function keeps grabbing the first element of ranges, each element of which is a two-element array, and passing those numbers into the range() method as two different values instead of one array. That's what the * does, it unpacks the array. First call, you unpack [1, 4] into range(1, 4) and then append the next call of the recursive function to it.

Basically this unpacks into the following:

list(range(1, 4)) + list(range(6, 11))

but you get to use a much more compact syntax, just passing a list of lists.

share|improve this answer

edited 9 hours ago

answered 9 hours ago

EngineeroEngineero

6,48632351

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  this will be too long for larger numbers
  
  – Led
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Led what do you mean? Also see my update with the recursive function. Pretty compact.
  
  – Engineero
  9 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks, but there is a need to write a function. Great idea by the way
  
  – Led
  9 hours ago
  

  
  
  
  

add a comment | 

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13

Simplest way to do this is to call range() and unpack result inside list assignment.

x = [*range(1, 4), *range(6, 11)]

share|improve this answer

edited 9 hours ago

answered 9 hours ago

Olvin RoghtOlvin Roght

34219

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Please name and refer to the docs of the corresponding syntax constructs
  
  – ivan_pozdeev
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  whats with the *?
  
  – Led
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Exactly this is the way to go. gj
  
  – Netwave
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @Led, Unpacking arguments
  
  – Olvin Roght
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  While this works, it's worth noting that this creates the entire list in memory. If you just want to iterate over the resulting sequence once, then this will consume more memory than the itertools.chain method. 40 bytes per number on my system, to be exact. This is fine for small sequences, but if your sequence has a billion numbers, well, you do the math...
  
  – marcelm
  6 hours ago
  

  
  
  
  

 | 
show 1 more comment

13

Simplest way to do this is to call range() and unpack result inside list assignment.

x = [*range(1, 4), *range(6, 11)]

share|improve this answer

edited 9 hours ago

answered 9 hours ago

Olvin RoghtOlvin Roght

34219

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Please name and refer to the docs of the corresponding syntax constructs
  
  – ivan_pozdeev
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  whats with the *?
  
  – Led
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Exactly this is the way to go. gj
  
  – Netwave
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @Led, Unpacking arguments
  
  – Olvin Roght
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  While this works, it's worth noting that this creates the entire list in memory. If you just want to iterate over the resulting sequence once, then this will consume more memory than the itertools.chain method. 40 bytes per number on my system, to be exact. This is fine for small sequences, but if your sequence has a billion numbers, well, you do the math...
  
  – marcelm
  6 hours ago
  

  
  
  
  

 | 
show 1 more comment

Simplest way to do this is to call range() and unpack result inside list assignment.

x = [*range(1, 4), *range(6, 11)]

share|improve this answer

edited 9 hours ago

answered 9 hours ago

Olvin RoghtOlvin Roght

34219

x = [*range(1, 4), *range(6, 11)]

share|improve this answer

edited 9 hours ago

answered 9 hours ago

Olvin RoghtOlvin Roght

34219

answered 9 hours ago

Olvin RoghtOlvin Roght

34219

answered 9 hours ago

Olvin RoghtOlvin Roght

34219

11

Alternatively you can use itertools.chain:

>>> import itertools

>>> list(itertools.chain(range(1, 5), range(20, 25)))

[1, 2, 3, 4, 20, 21, 22, 23, 24]

share|improve this answer

answered 9 hours ago

NetwaveNetwave

14.1k22247

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  interesting, but is there no native way? i mean I need to import something? thank you very much
  
  – Led
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  @Led, itertools is one of the most powerful libs that come with python, it is as native as it can be. Otherwise the unpacking sintax is handy also.
  
  – Netwave
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @Led why does it matter if you have to import from the standard library? It's part of Python
  
  – juanpa.arrivillaga
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  im looking for something short without importing.
  
  – Led
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Led Importing is your friend. Learn that sooner rather than later.
  
  – wizzwizz4
  6 hours ago
  

  
  
  
  

add a comment | 

11

Alternatively you can use itertools.chain:

>>> import itertools

>>> list(itertools.chain(range(1, 5), range(20, 25)))

[1, 2, 3, 4, 20, 21, 22, 23, 24]

share|improve this answer

answered 9 hours ago

NetwaveNetwave

14.1k22247

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  interesting, but is there no native way? i mean I need to import something? thank you very much
  
  – Led
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  @Led, itertools is one of the most powerful libs that come with python, it is as native as it can be. Otherwise the unpacking sintax is handy also.
  
  – Netwave
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @Led why does it matter if you have to import from the standard library? It's part of Python
  
  – juanpa.arrivillaga
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  im looking for something short without importing.
  
  – Led
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Led Importing is your friend. Learn that sooner rather than later.
  
  – wizzwizz4
  6 hours ago
  

  
  
  
  

add a comment | 

Alternatively you can use itertools.chain:

>>> import itertools

>>> list(itertools.chain(range(1, 5), range(20, 25)))

[1, 2, 3, 4, 20, 21, 22, 23, 24]

share|improve this answer

answered 9 hours ago

NetwaveNetwave

14.1k22247

>>> import itertools

>>> list(itertools.chain(range(1, 5), range(20, 25)))

[1, 2, 3, 4, 20, 21, 22, 23, 24]

share|improve this answer

answered 9 hours ago

NetwaveNetwave

14.1k22247

answered 9 hours ago

NetwaveNetwave

14.1k22247

answered 9 hours ago

NetwaveNetwave

14.1k22247

5

If numpy is an option, you can use np.r_ to concatenate slice objects:

import numpy as np

np.r_[range(1,4), range(6,11)]

# array([ 1,  2,  3,  6,  7,  8,  9, 10])

share|improve this answer

answered 9 hours ago

yatuyatu

15.8k41642

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  And since we just need slice objects you can write np.r_[1:4, 6:11] as well.
  
  – miradulo
  2 hours ago
  

  
  
  
  

add a comment | 

5

If numpy is an option, you can use np.r_ to concatenate slice objects:

import numpy as np

np.r_[range(1,4), range(6,11)]

# array([ 1,  2,  3,  6,  7,  8,  9, 10])

share|improve this answer

answered 9 hours ago

yatuyatu

15.8k41642

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  And since we just need slice objects you can write np.r_[1:4, 6:11] as well.
  
  – miradulo
  2 hours ago
  

  
  
  
  

add a comment | 

If numpy is an option, you can use np.r_ to concatenate slice objects:

import numpy as np

np.r_[range(1,4), range(6,11)]

# array([ 1,  2,  3,  6,  7,  8,  9, 10])

share|improve this answer

answered 9 hours ago

yatuyatu

15.8k41642

import numpy as np

np.r_[range(1,4), range(6,11)]

# array([ 1,  2,  3,  6,  7,  8,  9, 10])

share|improve this answer

answered 9 hours ago

yatuyatu

15.8k41642

answered 9 hours ago

yatuyatu

15.8k41642

answered 9 hours ago

yatuyatu

15.8k41642

2

You can turn it into a recursive function:

def recursive_ranges(ranges):

    if len(ranges) == 1:

        return list(range(*ranges[0]))

    else:

        return list(range(*ranges[0])) + recursive_ranges(ranges[1:])

You can then call this, specifying ranges as a list of lists:

ranges = [[1, 4], [6, 11]]

recursive_ranges(ranges)

# [1, 2, 3, 6, 7, 8, 9, 10]

Note the *ranges[0] is used to unpack the elements in ranges[0] into individual arguments. Essentially the recursive function keeps grabbing the first element of ranges, each element of which is a two-element array, and passing those numbers into the range() method as two different values instead of one array. That's what the * does, it unpacks the array. First call, you unpack [1, 4] into range(1, 4) and then append the next call of the recursive function to it.

Basically this unpacks into the following:

list(range(1, 4)) + list(range(6, 11))

but you get to use a much more compact syntax, just passing a list of lists.

share|improve this answer

edited 9 hours ago

answered 9 hours ago

EngineeroEngineero

6,48632351

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  this will be too long for larger numbers
  
  – Led
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Led what do you mean? Also see my update with the recursive function. Pretty compact.
  
  – Engineero
  9 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks, but there is a need to write a function. Great idea by the way
  
  – Led
  9 hours ago
  

  
  
  
  

add a comment | 

2

You can turn it into a recursive function:

def recursive_ranges(ranges):

    if len(ranges) == 1:

        return list(range(*ranges[0]))

    else:

        return list(range(*ranges[0])) + recursive_ranges(ranges[1:])

You can then call this, specifying ranges as a list of lists:

ranges = [[1, 4], [6, 11]]

recursive_ranges(ranges)

# [1, 2, 3, 6, 7, 8, 9, 10]

Note the *ranges[0] is used to unpack the elements in ranges[0] into individual arguments. Essentially the recursive function keeps grabbing the first element of ranges, each element of which is a two-element array, and passing those numbers into the range() method as two different values instead of one array. That's what the * does, it unpacks the array. First call, you unpack [1, 4] into range(1, 4) and then append the next call of the recursive function to it.

Basically this unpacks into the following:

list(range(1, 4)) + list(range(6, 11))

but you get to use a much more compact syntax, just passing a list of lists.

share|improve this answer

edited 9 hours ago

answered 9 hours ago

EngineeroEngineero

6,48632351

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  this will be too long for larger numbers
  
  – Led
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Led what do you mean? Also see my update with the recursive function. Pretty compact.
  
  – Engineero
  9 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks, but there is a need to write a function. Great idea by the way
  
  – Led
  9 hours ago
  

  
  
  
  

add a comment | 

You can turn it into a recursive function:

def recursive_ranges(ranges):

    if len(ranges) == 1:

        return list(range(*ranges[0]))

    else:

        return list(range(*ranges[0])) + recursive_ranges(ranges[1:])

You can then call this, specifying ranges as a list of lists:

ranges = [[1, 4], [6, 11]]

recursive_ranges(ranges)

# [1, 2, 3, 6, 7, 8, 9, 10]

Note the *ranges[0] is used to unpack the elements in ranges[0] into individual arguments. Essentially the recursive function keeps grabbing the first element of ranges, each element of which is a two-element array, and passing those numbers into the range() method as two different values instead of one array. That's what the * does, it unpacks the array. First call, you unpack [1, 4] into range(1, 4) and then append the next call of the recursive function to it.

Basically this unpacks into the following:

list(range(1, 4)) + list(range(6, 11))

but you get to use a much more compact syntax, just passing a list of lists.

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edited 9 hours ago

answered 9 hours ago

EngineeroEngineero

6,48632351

def recursive_ranges(ranges):

    if len(ranges) == 1:

        return list(range(*ranges[0]))

    else:

        return list(range(*ranges[0])) + recursive_ranges(ranges[1:])

ranges = [[1, 4], [6, 11]]

recursive_ranges(ranges)

# [1, 2, 3, 6, 7, 8, 9, 10]

list(range(1, 4)) + list(range(6, 11))

share|improve this answer

edited 9 hours ago

answered 9 hours ago

EngineeroEngineero

6,48632351

answered 9 hours ago

EngineeroEngineero

6,48632351

answered 9 hours ago

EngineeroEngineero

6,48632351