Convert a list into chunks of increasing size

-1

How could I convert a list into a nested list with increasing size of the sublists?

For example,
from

[1, 2, 3, 4, 5, 6]

[[1], [2, 3], [4, 5, 6]]

edited Nov 18 '18 at 20:01

timgeb

50.4k116391

asked Nov 18 '18 at 19:41

Ivan

2

that supposes that the length of the list is of the form (n+1)*n // 2 for starters. Is there something you tried?
– Jean-François Fabre
Nov 18 '18 at 19:43

And what have you tried so far?
– schwobaseggl
Nov 18 '18 at 19:43

1

By writing some code. How do you decide the length of each sub-array? Would a list [2,3,4] get split in [[2,3],[4]]? If the sequence of lengths is 1,2,3... what happens with the last element if these do not add up?
– usr2564301
Nov 18 '18 at 19:45

this could help (with adaptation): stackoverflow.com/questions/312443/…
– Jean-François Fabre
Nov 18 '18 at 19:45

Welcome to SO! It's unclear how odd-length lists should be handled. Please clarify.
– ggorlen
Nov 18 '18 at 20:06

add a comment |

-1

How could I convert a list into a nested list with increasing size of the sublists?

For example,
from

[1, 2, 3, 4, 5, 6]

[[1], [2, 3], [4, 5, 6]]

edited Nov 18 '18 at 20:01

timgeb

50.4k116391

asked Nov 18 '18 at 19:41

Ivan

2

that supposes that the length of the list is of the form (n+1)*n // 2 for starters. Is there something you tried?
– Jean-François Fabre
Nov 18 '18 at 19:43

And what have you tried so far?
– schwobaseggl
Nov 18 '18 at 19:43

1

By writing some code. How do you decide the length of each sub-array? Would a list [2,3,4] get split in [[2,3],[4]]? If the sequence of lengths is 1,2,3... what happens with the last element if these do not add up?
– usr2564301
Nov 18 '18 at 19:45

this could help (with adaptation): stackoverflow.com/questions/312443/…
– Jean-François Fabre
Nov 18 '18 at 19:45

Welcome to SO! It's unclear how odd-length lists should be handled. Please clarify.
– ggorlen
Nov 18 '18 at 20:06

add a comment |

-1

How could I convert a list into a nested list with increasing size of the sublists?

For example,
from

[1, 2, 3, 4, 5, 6]

[[1], [2, 3], [4, 5, 6]]

edited Nov 18 '18 at 20:01

timgeb

50.4k116391

asked Nov 18 '18 at 19:41

Ivan

How could I convert a list into a nested list with increasing size of the sublists?

For example,
from

[1, 2, 3, 4, 5, 6]

[[1], [2, 3], [4, 5, 6]]

python list

edited Nov 18 '18 at 20:01

timgeb

50.4k116391

asked Nov 18 '18 at 19:41

Ivan

edited Nov 18 '18 at 20:01

timgeb

50.4k116391

asked Nov 18 '18 at 19:41

Ivan

edited Nov 18 '18 at 20:01

timgeb

50.4k116391

edited Nov 18 '18 at 20:01

timgeb

50.4k116391

edited Nov 18 '18 at 20:01

timgeb

50.4k116391

asked Nov 18 '18 at 19:41

Ivan

asked Nov 18 '18 at 19:41

Ivan

asked Nov 18 '18 at 19:41

Ivan

2

that supposes that the length of the list is of the form (n+1)*n // 2 for starters. Is there something you tried?
– Jean-François Fabre
Nov 18 '18 at 19:43

And what have you tried so far?
– schwobaseggl
Nov 18 '18 at 19:43

1

By writing some code. How do you decide the length of each sub-array? Would a list [2,3,4] get split in [[2,3],[4]]? If the sequence of lengths is 1,2,3... what happens with the last element if these do not add up?
– usr2564301
Nov 18 '18 at 19:45

this could help (with adaptation): stackoverflow.com/questions/312443/…
– Jean-François Fabre
Nov 18 '18 at 19:45

Welcome to SO! It's unclear how odd-length lists should be handled. Please clarify.
– ggorlen
Nov 18 '18 at 20:06

add a comment |

2

that supposes that the length of the list is of the form (n+1)*n // 2 for starters. Is there something you tried?
– Jean-François Fabre
Nov 18 '18 at 19:43

And what have you tried so far?
– schwobaseggl
Nov 18 '18 at 19:43

1

By writing some code. How do you decide the length of each sub-array? Would a list [2,3,4] get split in [[2,3],[4]]? If the sequence of lengths is 1,2,3... what happens with the last element if these do not add up?
– usr2564301
Nov 18 '18 at 19:45

this could help (with adaptation): stackoverflow.com/questions/312443/…
– Jean-François Fabre
Nov 18 '18 at 19:45

Welcome to SO! It's unclear how odd-length lists should be handled. Please clarify.
– ggorlen
Nov 18 '18 at 20:06

that supposes that the length of the list is of the form (n+1)*n // 2 for starters. Is there something you tried?
– Jean-François Fabre
Nov 18 '18 at 19:43

And what have you tried so far?
– schwobaseggl
Nov 18 '18 at 19:43

By writing some code. How do you decide the length of each sub-array? Would a list [2,3,4] get split in [[2,3],[4]]? If the sequence of lengths is 1,2,3... what happens with the last element if these do not add up?
– usr2564301
Nov 18 '18 at 19:45

this could help (with adaptation): stackoverflow.com/questions/312443/…
– Jean-François Fabre
Nov 18 '18 at 19:45

Welcome to SO! It's unclear how odd-length lists should be handled. Please clarify.
– ggorlen
Nov 18 '18 at 20:06

add a comment |

4 Answers
4

active

oldest

votes

I'd do this with islices over an iterator of the original list. This way I can just specifiy the number of elements to take without having to worry at which position I am currently at. (In addition, the following code works with any iterable.)

def increasing_chunks(iterable):

    it = iter(iterable)

    i = 1



    while True:

        chunk = list(islice(it, i))

        if not chunk:

            break

        yield chunk

        i += 1

The last chunk might be truncated to whatever amount of elements the iterator had left.

Demo:

>>> list(increasing_chunks([1, 2, 3, 4, 5, 6]))

[[1], [2, 3], [4, 5, 6]]

>>> list(increasing_chunks([1, 2, 3, 4, 5, 6, 7, 8]))

[[1], [2, 3], [4, 5, 6], [7, 8]]

If you want to discard truncated chunks, adjust the code as follows:

def increasing_chunks_strict(iterable):

    it = iter(iterable)

    i = 1



    while True:

        chunk = list(islice(it, i))

        if len(chunk) < i:

            break

        yield chunk

        i += 1

Now, truncated chunks are not included in the result.

>>> list(increasing_chunks_strict([1, 2, 3, 4, 5, 6]))

[[1], [2, 3], [4, 5, 6]]

>>> list(increasing_chunks_strict([1, 2, 3, 4, 5, 6, 7, 8]))

[[1], [2, 3], [4, 5, 6]]

edited Nov 18 '18 at 20:10

answered Nov 18 '18 at 19:48

timgeb

50.4k116391

add a comment |

As a follow up to timgeb's solution, without itertools, you need to keep track of the index:

l = [1, 2, 3, 4, 5, 6]



i, slice_length = 0, 1

result = 

while i < len(l):

    result.append(l[i:i + slice_length])

    i += slice_length

    slice_length += 1



print(result)

# [[1], [2, 3], [4, 5, 6]]

answered Nov 18 '18 at 20:12

slider

8,10011129

add a comment |

Several itertools and enumerate to the rescue:

from itertools import count, accumulate as acc, takewhile as tw



lst = [1, 2, 3, 4, 5, 6]

[lst[c:c+i] for i, c in enumerate(tw(lambda x: x < len(lst), acc(count())), 1)]

# [[1], [2, 3], [4, 5, 6]]

answered Nov 18 '18 at 20:01

schwobaseggl

36.8k32442

add a comment |

Assuming that you list length has the correct length for the last chunk to have the correct size, you can use list sum, range and list comprehension to solve your problem in few lines:

l = [1, 2, 3, 4, 5, 6]

slices = range(1, (len(l) + 1)/2 + 1)

result = [l[sum(slices[:s-1]):sum(slices[:s-1])+s] for s in slices]

answered Nov 18 '18 at 20:07

Aurora Wang

715216

add a comment |

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

def increasing_chunks(iterable):

    it = iter(iterable)

    i = 1



    while True:

        chunk = list(islice(it, i))

        if not chunk:

            break

        yield chunk

        i += 1

The last chunk might be truncated to whatever amount of elements the iterator had left.

Demo:

>>> list(increasing_chunks([1, 2, 3, 4, 5, 6]))

[[1], [2, 3], [4, 5, 6]]

>>> list(increasing_chunks([1, 2, 3, 4, 5, 6, 7, 8]))

[[1], [2, 3], [4, 5, 6], [7, 8]]

If you want to discard truncated chunks, adjust the code as follows:

def increasing_chunks_strict(iterable):

    it = iter(iterable)

    i = 1



    while True:

        chunk = list(islice(it, i))

        if len(chunk) < i:

            break

        yield chunk

        i += 1

Now, truncated chunks are not included in the result.

>>> list(increasing_chunks_strict([1, 2, 3, 4, 5, 6]))

[[1], [2, 3], [4, 5, 6]]

>>> list(increasing_chunks_strict([1, 2, 3, 4, 5, 6, 7, 8]))

[[1], [2, 3], [4, 5, 6]]

edited Nov 18 '18 at 20:10

answered Nov 18 '18 at 19:48

timgeb

50.4k116391

add a comment |

def increasing_chunks(iterable):

    it = iter(iterable)

    i = 1



    while True:

        chunk = list(islice(it, i))

        if not chunk:

            break

        yield chunk

        i += 1

The last chunk might be truncated to whatever amount of elements the iterator had left.

Demo:

>>> list(increasing_chunks([1, 2, 3, 4, 5, 6]))

[[1], [2, 3], [4, 5, 6]]

>>> list(increasing_chunks([1, 2, 3, 4, 5, 6, 7, 8]))

[[1], [2, 3], [4, 5, 6], [7, 8]]

If you want to discard truncated chunks, adjust the code as follows:

def increasing_chunks_strict(iterable):

    it = iter(iterable)

    i = 1



    while True:

        chunk = list(islice(it, i))

        if len(chunk) < i:

            break

        yield chunk

        i += 1

Now, truncated chunks are not included in the result.

>>> list(increasing_chunks_strict([1, 2, 3, 4, 5, 6]))

[[1], [2, 3], [4, 5, 6]]

>>> list(increasing_chunks_strict([1, 2, 3, 4, 5, 6, 7, 8]))

[[1], [2, 3], [4, 5, 6]]

edited Nov 18 '18 at 20:10

answered Nov 18 '18 at 19:48

timgeb

50.4k116391

add a comment |

def increasing_chunks(iterable):

    it = iter(iterable)

    i = 1



    while True:

        chunk = list(islice(it, i))

        if not chunk:

            break

        yield chunk

        i += 1

The last chunk might be truncated to whatever amount of elements the iterator had left.

Demo:

>>> list(increasing_chunks([1, 2, 3, 4, 5, 6]))

[[1], [2, 3], [4, 5, 6]]

>>> list(increasing_chunks([1, 2, 3, 4, 5, 6, 7, 8]))

[[1], [2, 3], [4, 5, 6], [7, 8]]

If you want to discard truncated chunks, adjust the code as follows:

def increasing_chunks_strict(iterable):

    it = iter(iterable)

    i = 1



    while True:

        chunk = list(islice(it, i))

        if len(chunk) < i:

            break

        yield chunk

        i += 1

Now, truncated chunks are not included in the result.

>>> list(increasing_chunks_strict([1, 2, 3, 4, 5, 6]))

[[1], [2, 3], [4, 5, 6]]

>>> list(increasing_chunks_strict([1, 2, 3, 4, 5, 6, 7, 8]))

[[1], [2, 3], [4, 5, 6]]

edited Nov 18 '18 at 20:10

answered Nov 18 '18 at 19:48

timgeb

50.4k116391

def increasing_chunks(iterable):

    it = iter(iterable)

    i = 1



    while True:

        chunk = list(islice(it, i))

        if not chunk:

            break

        yield chunk

        i += 1

The last chunk might be truncated to whatever amount of elements the iterator had left.

Demo:

>>> list(increasing_chunks([1, 2, 3, 4, 5, 6]))

[[1], [2, 3], [4, 5, 6]]

>>> list(increasing_chunks([1, 2, 3, 4, 5, 6, 7, 8]))

[[1], [2, 3], [4, 5, 6], [7, 8]]

If you want to discard truncated chunks, adjust the code as follows:

def increasing_chunks_strict(iterable):

    it = iter(iterable)

    i = 1



    while True:

        chunk = list(islice(it, i))

        if len(chunk) < i:

            break

        yield chunk

        i += 1

Now, truncated chunks are not included in the result.

>>> list(increasing_chunks_strict([1, 2, 3, 4, 5, 6]))

[[1], [2, 3], [4, 5, 6]]

>>> list(increasing_chunks_strict([1, 2, 3, 4, 5, 6, 7, 8]))

[[1], [2, 3], [4, 5, 6]]

edited Nov 18 '18 at 20:10

answered Nov 18 '18 at 19:48

timgeb

50.4k116391

edited Nov 18 '18 at 20:10

answered Nov 18 '18 at 19:48

timgeb

50.4k116391

answered Nov 18 '18 at 19:48

timgeb

50.4k116391

answered Nov 18 '18 at 19:48

timgeb

50.4k116391

add a comment |

As a follow up to timgeb's solution, without itertools, you need to keep track of the index:

l = [1, 2, 3, 4, 5, 6]



i, slice_length = 0, 1

result = 

while i < len(l):

    result.append(l[i:i + slice_length])

    i += slice_length

    slice_length += 1



print(result)

# [[1], [2, 3], [4, 5, 6]]

answered Nov 18 '18 at 20:12

slider

8,10011129

add a comment |

As a follow up to timgeb's solution, without itertools, you need to keep track of the index:

l = [1, 2, 3, 4, 5, 6]



i, slice_length = 0, 1

result = 

while i < len(l):

    result.append(l[i:i + slice_length])

    i += slice_length

    slice_length += 1



print(result)

# [[1], [2, 3], [4, 5, 6]]

answered Nov 18 '18 at 20:12

slider

8,10011129

add a comment |

As a follow up to timgeb's solution, without itertools, you need to keep track of the index:

l = [1, 2, 3, 4, 5, 6]



i, slice_length = 0, 1

result = 

while i < len(l):

    result.append(l[i:i + slice_length])

    i += slice_length

    slice_length += 1



print(result)

# [[1], [2, 3], [4, 5, 6]]

answered Nov 18 '18 at 20:12

slider

8,10011129

As a follow up to timgeb's solution, without itertools, you need to keep track of the index:

l = [1, 2, 3, 4, 5, 6]



i, slice_length = 0, 1

result = 

while i < len(l):

    result.append(l[i:i + slice_length])

    i += slice_length

    slice_length += 1



print(result)

# [[1], [2, 3], [4, 5, 6]]

answered Nov 18 '18 at 20:12

slider

8,10011129

answered Nov 18 '18 at 20:12

slider

8,10011129

answered Nov 18 '18 at 20:12

slider

8,10011129

answered Nov 18 '18 at 20:12

slider

8,10011129

add a comment |

Several itertools and enumerate to the rescue:

from itertools import count, accumulate as acc, takewhile as tw



lst = [1, 2, 3, 4, 5, 6]

[lst[c:c+i] for i, c in enumerate(tw(lambda x: x < len(lst), acc(count())), 1)]

# [[1], [2, 3], [4, 5, 6]]

answered Nov 18 '18 at 20:01

schwobaseggl

36.8k32442

add a comment |

Several itertools and enumerate to the rescue:

from itertools import count, accumulate as acc, takewhile as tw



lst = [1, 2, 3, 4, 5, 6]

[lst[c:c+i] for i, c in enumerate(tw(lambda x: x < len(lst), acc(count())), 1)]

# [[1], [2, 3], [4, 5, 6]]

answered Nov 18 '18 at 20:01

schwobaseggl

36.8k32442

add a comment |

Several itertools and enumerate to the rescue:

from itertools import count, accumulate as acc, takewhile as tw



lst = [1, 2, 3, 4, 5, 6]

[lst[c:c+i] for i, c in enumerate(tw(lambda x: x < len(lst), acc(count())), 1)]

# [[1], [2, 3], [4, 5, 6]]

answered Nov 18 '18 at 20:01

schwobaseggl

36.8k32442

Several itertools and enumerate to the rescue:

from itertools import count, accumulate as acc, takewhile as tw



lst = [1, 2, 3, 4, 5, 6]

[lst[c:c+i] for i, c in enumerate(tw(lambda x: x < len(lst), acc(count())), 1)]

# [[1], [2, 3], [4, 5, 6]]

answered Nov 18 '18 at 20:01

schwobaseggl

36.8k32442

answered Nov 18 '18 at 20:01

schwobaseggl

36.8k32442

answered Nov 18 '18 at 20:01

schwobaseggl

36.8k32442

answered Nov 18 '18 at 20:01

schwobaseggl

36.8k32442

add a comment |

Assuming that you list length has the correct length for the last chunk to have the correct size, you can use list sum, range and list comprehension to solve your problem in few lines:

l = [1, 2, 3, 4, 5, 6]

slices = range(1, (len(l) + 1)/2 + 1)

result = [l[sum(slices[:s-1]):sum(slices[:s-1])+s] for s in slices]

answered Nov 18 '18 at 20:07

Aurora Wang

715216

add a comment |

Assuming that you list length has the correct length for the last chunk to have the correct size, you can use list sum, range and list comprehension to solve your problem in few lines:

l = [1, 2, 3, 4, 5, 6]

slices = range(1, (len(l) + 1)/2 + 1)

result = [l[sum(slices[:s-1]):sum(slices[:s-1])+s] for s in slices]

answered Nov 18 '18 at 20:07

Aurora Wang

715216

add a comment |

Assuming that you list length has the correct length for the last chunk to have the correct size, you can use list sum, range and list comprehension to solve your problem in few lines:

l = [1, 2, 3, 4, 5, 6]

slices = range(1, (len(l) + 1)/2 + 1)

result = [l[sum(slices[:s-1]):sum(slices[:s-1])+s] for s in slices]

answered Nov 18 '18 at 20:07

Aurora Wang

715216

Assuming that you list length has the correct length for the last chunk to have the correct size, you can use list sum, range and list comprehension to solve your problem in few lines:

l = [1, 2, 3, 4, 5, 6]

slices = range(1, (len(l) + 1)/2 + 1)

result = [l[sum(slices[:s-1]):sum(slices[:s-1])+s] for s in slices]

answered Nov 18 '18 at 20:07

Aurora Wang

715216

answered Nov 18 '18 at 20:07

Aurora Wang

715216

answered Nov 18 '18 at 20:07

Aurora Wang

715216

answered Nov 18 '18 at 20:07

Aurora Wang

715216

add a comment |

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