## How to reshape python list without numpy

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How do I reshape a list into a n-dimensional list

Input: list = `[1, 2, 3, 4, 5, 6, 7, 8]` shape = `[2, 2, 2]`

output = `[[[1, 2], [3, 4]], [[5, 6], [7, 8]]]`

This recursive approach should work.

```lst = [1, 2, 3, 4, 5, 6, 7, 8]
shape = [2, 2, 2]

from functools import reduce
from operator import mul

def reshape(lst, shape):
if len(shape) == 1:
return lst
n = reduce(mul, shape[1:])
return [reshape(lst[i*n:(i+1)*n], shape[1:]) for i in range(len(lst)//n)]

reshape(lst, shape)
```

You probably want to wrap that with a check that your dimensions make sense... e.g.

```assert reduce(mul, shape) == len(lst)
```

[Python] with and without Numpy, First check dimension conditions, then reshape with numpy. with a list comprehension, then build a new matrix with a list comprehension. numpy.reshape() function. The reshape() function is used to give a new shape to an array without changing its data. Syntax: numpy.reshape(a, newshape, order='C')

Here is an approach using the grouper once on each dimension except the first:

```import functools as ft

# example
L = list(range(2*3*4))
S = 2,3,4

# if tuples are acceptable
tuple(ft.reduce(lambda x, y: zip(*y*(x,)), (iter(L), *S[:0:-1])))
# (((0, 1, 2, 3), (4, 5, 6, 7), (8, 9, 10, 11)), ((12, 13, 14, 15), (16, 17, 18, 19), (20, 21, 22, 23)))

# if it must be lists
list(ft.reduce(lambda x, y: map(list, zip(*y*(x,))), (iter(L), *S[:0:-1])))
# [[[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11]], [[12, 13, 14, 15], [16, 17, 18, 19], [20, 21, 22, 23]]]
```

Creating 2D array without Numpy, So you have a list of references, not a list of lists. The second way a new [0] * n is created each time through the loop. That way there is no� numpy.reshape¶ numpy.reshape (a, newshape, order='C') [source] ¶ Gives a new shape to an array without changing its data. Parameters a array_like. Array to be reshaped. newshape int or tuple of ints. The new shape should be compatible with the original shape. If an integer, then the result will be a 1-D array of that length. One shape

Not particularly elegant:

```from functools import reduce
from itertools import islice
l=[1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9,0,1,2,3,4]
s=[2,3,4]
if s and reduce(lambda x,y:x*y, s) == len(l):
# if number of elements matches product of dimensions,
# the first dimension is actually redundant
s=[1:]
else:
print("length of input list does not match shape")
return
while s:
size = s.pop()  # how many elements for this dimension
#split the list based on the size of the dimension
it=iter(l)
l = list(iter(lambda:list(islice(it,size)),[]))

# [[[1, 2, 3, 4], [5, 6, 7, 8], [9, 0, 1, 2]],
#  [[3, 4, 5, 6], [7, 8, 9, 0], [1, 2, 3, 4]]]
```

Shape and Reshape, Using the shape and reshape tools available in the NumPy module, configure a list according to the guidelines. The reshape tool gives a new shape to an array without changing its data. It creates a You are given a space separated list of nine integers. Python 2. 1. 2. 3. 4. 5. import numpy. import numpy. Line: 5 Col: 1. In this article, you will learn, How to reshape numpy arrays in python using numpy.reshape() function. Before going further into article, first learn about numpy.reshape() function syntax and it’s parameters. Syntax: numpy.reshape(a, newshape, order=’C’) This function helps to get a new shape to an array without changing its data. Parameters:

The code below should do the trick.

The solution given below very general. The input list can be a nested list of lists of an any old/undesired shape; it need not be a list of integers.

Also, there are separate re-usable tools. For example the `all_for_one` function is very handy.

EDIT: I failed to note something important. If you put `1`s inside of the `shape` parameter, then you can get superfluous list nestings (only one list inside of a list instead of five or six lists inside of a list)

For example, if shape is `[1, 1, 2]` then the return value might be `[[[0.1, 0.2]]]` instead of `[0.1, 0.2]`

the length of `shape` is the number of valid subscripts in the output list. For example,

```shape = [1, 2] # length 2
lyst = [[0.1, 0.2]]
print(lyst[0][0])  # valid.... no KeyError raised
```

If you want a true column or row vector, then `len(shape)` must be `1`. For example, `shape = [49]` will give you a row/column vector of length `49`.

```shape = [2] # length 2
output = [0.1, 0.2]
print(lyst[0])
```

Here's the code:

```from operator import mul
import itertools as itts
import copy
import functools

one_for_all =  lambda one: itts.repeat(one, 1)

def all_for_one(lyst):
"""
EXAMPLE:
INPUT:
[[[1, 2], [3, 4]], [[5, 6], [7, 8]]]
OUTPUT:
iterator to [1, 2, 3, 4, 5, 6, 7, 8]

IN GENERAL:
Gets iterator to all nested elements
of a list of lists of ... of lists of lists.
"""
# make an iterator which **IMMEDIATELY**
# raises a `StopIteration` exception
its = itts.repeat("", 0)

for sublyst in lyst:

if hasattr(sublyst, "__iter__") and id(sublyst) != id(lyst):
# Be careful ....
#
# "string"[0] == "s"[0] == "s"[0][0][0][0][0][0]...
#
# do not drill down while `sublyst` has an "__iter__" method
# do not drill down while `sublyst` has a `__getitem__` method
#
it = all_for_one(sublyst)
else:
it = one_for_all(sublyst)

# concatenate results to what we had previously
its = itts.chain(its, it)

return its

merged = list(all_for_one([[[1, 2], [3, 4]], [[5, 6], [7, 8]]]))
print("merged == ", merged)

"""
similar to `numpy.reshape`

EXAMPLE:

lyst  = [1, 2, 3, 4, 5, 6, 7, 8]
shape = [2, 2, 2]
result = reshape(lyst)
print(result)

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

For this function, input parameter `xshape` can be
any iterable containing at least one element.
`xshape` is not required to be a tuple, but it can be.

The length of xshape should be equal to the number
of desired list nestings

If you want a list of integers: len(xshape) == 1
If you want a list of lists:    len(xshape) == 2
If you want a list of lists of lists: len(xshape) == 3

If xshape = [1, 2],
outermost list has 1 element
that one element is a list of 2 elements.
result == [[1, 2]]

If xshape == [2]
outermost list has 2 elements
those 2 elements are non-lists:
result: [1, 2]

If xshape = [2, 2],
outermost list has 2 elements
each element is a list of 2 elements.
result == [[1, 2] [3, 4]]

"""
# BEGIN SANITIZING INPUTS

# unfortunately, iterators are not re-usable
# Also, they don't have `len` methods

ishape = [x for x in self.unnest(xshape)]

number_of_elements = functools.reduce(mul, ishape, 1)

msg = [str(x) for x in [
"\nAn array having dimensions ", ishape,
"\nMust contain ", number_of_elements, " element(s).",
"\nHowever, we were only given ", len(iread_lyst), " element(s)."
]]
msg.append('\nList before reshape: ')
raise TypeError(''.join(msg))
ishape = iter(ishape)
# END SANITATIZATION OF INPUTS
write_parent = list()
parent_list_len = next(ishape)
try:
child_list_len = next(ishape)
for _ in range(0, parent_list_len):
write_child = []
write_parent.append(write_child)
except StopIteration:
for _ in range(0, parent_list_len):
write_parent.append(write_child)
return write_parent

"""
You really shouldn't call this function directly.

The `i` in the name of this function stands for "internal"
"""
try:
child_list_len = next(ishape)
for _ in range(0, parent_list_len):
write_child = []
write_parent.append(write_child)
except StopIteration:
for _ in range(0, parent_list_len):
write_parent.append(write_child)
return None

three_dee_mat = reshape(merged, [2, 2, 2])
print("three_dee_mat == ", three_dee_mat)
```

Python, Given two lists, a single dimensional and a multidimensional list, write Python program to reshape the single dimensional list according to the� Reshape Data. In some occasions, you need to reshape the data from wide to long. You can use the reshape function for this. The syntax is numpy.reshape(a, newShape, order='C') Here, a: Array that you want to reshape . newShape: The new desires shape . Order: Default is C which is an essential row style. Exampe of Reshape

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numpy.reshape() in Python. The numpy.reshape() function is available in NumPy package. As the name suggests, reshape means 'changes in shape'. The numpy.reshape() function helps us to get a new shape to an array without changing its data. Sometimes, we need to reshape the data from wide to long.

In python, reshaping numpy array can be very critical while creating a matrix or tensor from vectors. In order to reshape numpy array of one dimension to n dimensions one can use np.reshape() method. Let’s check out some simple examples. It is very important to reshape you numpy array, especially you are training with some deep learning network.

• I am wondering if there is a way to use this call in Python 2.7? I ran this command, but it appears that the `*S` term is not recognized in python 2.7: `SyntaxError: invalid syntax`.
• @FangQ You can add tuples instead: `(a,*B)` in py3 can be replaced with `(a,) + B` in py2 if `B` is a tuple.