create 2d array python for loop

Create 2d Array in Python Using For Loop Results [duplicate]

stackoverflow.com › questions › 35403200 › create-2d-array-in-python-using-for-loop-results

You can try this:

import numpy as np
listvals = []
for i in range(10):
    listvals.append((xc, yc))
mat_vals = np.vstack(listvals)

This will output an ndarray like this:

[[ 2  0]
 [ 3  1]
 [ 4  2]
 [ 5  3]
 [ 6  4]
 [ 7  5]
 [ 8  6]
 [ 9  7]
 [10  8]
 [11  9]]

Or this maybe is better:

import numpy as np
list_xc = []
list_yc = []
for i in np.arange(10): 
    list_xc.append(xc)
    list_yc.append(yc)
mat_xc = np.asarray(list_xc)
mat_yc = np.asarray(list_yc)
mat_f = np.column_stack((mat_xc, mat_yc))

Answer from Litwos on Stack Overflow

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stackoverflow.com › questions › 35403200 › create-2d-array-in-python-using-for-loop-results

Create 2d Array in Python Using For Loop Results - Stack Overflow

Top answer

1 of 3

You can try this:

import numpy as np
listvals = []
for i in range(10):
    listvals.append((xc, yc))
mat_vals = np.vstack(listvals)

This will output an ndarray like this:

[[ 2  0]
 [ 3  1]
 [ 4  2]
 [ 5  3]
 [ 6  4]
 [ 7  5]
 [ 8  6]
 [ 9  7]
 [10  8]
 [11  9]]

Or this maybe is better:

import numpy as np
list_xc = []
list_yc = []
for i in np.arange(10): 
    list_xc.append(xc)
    list_yc.append(yc)
mat_xc = np.asarray(list_xc)
mat_yc = np.asarray(list_yc)
mat_f = np.column_stack((mat_xc, mat_yc))

2 of 3

You can do this:

aList = []
for i in range(10):
    aList.append([xc, yc])

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To process 2-dimensional array, you typically use nested loops. The first loop iterates through the row number, the second loop runs through the elements inside of a row.

Discussions

How can I create multiple arrays using a for loop or to convert into a 2D array

If you are using numpy and want output like this: array1 = [1, 2, 3] array2 = [3, 2, 1] result = [[3, 2, 1], [6, 4, 2], [9, 6, 3]] you can use np.outer(): np.outer(array1, array2) Edit: Not sure why this was downvoted, but this produces the output u/maybethisonework wants according to their comment above. np.outer([5, 1, 6, 2], [1, 5, 3, 9]) ## [[5,25,15,45], [1,5,3,9], ...] More on reddit.com

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How to initialize a two-dimensional array (list of lists, if not using NumPy) in Python? - Stack Overflow

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Those loops set in a 2D array each time value. value can be in this case whatever you want it to be. The first index determines the array in the array and the second index selects the value in the nested array. In your case when you have [[1, 2, 3], [4, 5, 6], [7, 8, 9]], array[1][2] == 6 · This also works for arrays that are not symmetric. ... Oups! Lists are mutable objects in Python... More on stackoverflow.com

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Top answer

1 of 5

2 of 5

In your new array, you're multiplying the list with a number. Make sure you put array2[i] in your for loop when you're adding to your new array. So, new_array = array1[i] * array2[i]. Update: misread question. You can loop through with your for loop, then newArray.append([]), then you can loop through your first array and multiply the value in the second array and add the result to newArray[i]. Hope it makes sense, sorry I'm on mobile.

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stackoverflow.com › questions › 2397141 › how-to-initialize-a-two-dimensional-array-list-of-lists-if-not-using-numpy-in

How to initialize a two-dimensional array (list of lists, if not using NumPy) in Python? - Stack Overflow

Top answer

1 of 16

652

To initialize a two-dimensional list in Python, use

t = [ [0]*3 for i in range(3)]

But don't use [[v]*n]*n, it is a trap!

>>> a = [[0]*3]*3
>>> a
[[0, 0, 0], [0, 0, 0], [0, 0, 0]]
>>> a[0][0]=1
>>> a
[[1, 0, 0], [1, 0, 0], [1, 0, 0]]

2 of 16

420

A pattern that often came up in Python was

bar = []
for item in some_iterable:
    bar.append(SOME EXPRESSION)

which helped motivate the introduction of list comprehensions, which convert that snippet to

bar = [SOME_EXPRESSION for item in some_iterable]

which is shorter and sometimes clearer. Usually, you get in the habit of recognizing these and often replacing loops with comprehensions.

Your code follows this pattern twice

twod_list = []                                       \                      
for i in range (0, 10):                               \
    new = []                  \ can be replaced        } this too
    for j in range (0, 10):    } with a list          /
        new.append(foo)       / comprehension        /
    twod_list.append(new)                           /

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stackoverflow.com › questions › 45707790 › creating-2d-arrays-and-input-values-in-it-with-a-while-loop

python - Creating 2D arrays and input values in it with a while loop? - Stack Overflow

Top answer

1 of 2

To go over a 2D array you have to use two nested loops.

i = 0
j = 0
while (i < len(array)):
    while(j < len(array[i]):
        array[i][j] = value
        j = j + 1
    j = 0
    i = i + 1

This also works for arrays that are not symmetric.

2 of 2

Oups! Lists are mutable objects in Python, and a list contains reference to objects. So when you write array = [[0]* days]*city, the external list contains city pointers to an unique list initialized with days 0.

Demo:

>>> array = [[0]* 3]*3
>>> array[0][0] = 1
>>> array
[[1, 0, 0], [1, 0, 0], [1, 0, 0]]

You should write instead array = [ [0] * 3 for i in range(3)]

Demo:

>>> array = [ [0] * 3 for i in range(3)]
>>> array[0][0] = 1
>>> array
[[1, 0, 0], [0, 0, 0], [0, 0, 0]]

For your question just cascade loops:

for c in range(city):
    for day in range(days):
        array[c][day] = int(input("temperature: "))

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Top answer

1 of 7

1263

You're technically trying to index an uninitialized array. You have to first initialize the outer list with lists before adding items; Python calls this "list comprehension".

# Creates a list containing 5 lists, each of 8 items, all set to 0
w, h = 8, 5
Matrix = [[0 for x in range(w)] for y in range(h)]

#You can now add items to the list:

Matrix[0][0] = 1
Matrix[6][0] = 3 # error! range... 
Matrix[0][6] = 3 # valid

Note that the matrix is "y" address major, in other words, the "y index" comes before the "x index".

print Matrix[0][0] # prints 1
x, y = 0, 6 
print Matrix[x][y] # prints 3; be careful with indexing!

Although you can name them as you wish, I look at it this way to avoid some confusion that could arise with the indexing, if you use "x" for both the inner and outer lists, and want a non-square Matrix.

2 of 7

487

If you really want a matrix, you might be better off using numpy. Matrix operations in numpy most often use an array type with two dimensions. There are many ways to create a new array; one of the most useful is the zeros function, which takes a shape parameter and returns an array of the given shape, with the values initialized to zero:

>>> import numpy
>>> numpy.zeros((5, 5))
array([[ 0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.]])

Here are some other ways to create 2-d arrays and matrices (with output removed for compactness):

numpy.arange(25).reshape((5, 5))         # create a 1-d range and reshape
numpy.array(range(25)).reshape((5, 5))   # pass a Python range and reshape
numpy.array([5] * 25).reshape((5, 5))    # pass a Python list and reshape
numpy.empty((5, 5))                      # allocate, but don't initialize
numpy.ones((5, 5))                       # initialize with ones

numpy provides a matrix type as well, but it is no longer recommended for any use, and may be removed from numpy in the future.

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stackoverflow.com › questions › 7536465 › create-a-2d-array-with-a-nested-loop

python - Create a 2D array with a nested loop - Stack Overflow

Top answer

1 of 3

Note this:

>>> matrix = [[0] * 3] * 3
>>> [x for x in matrix]
[[0, 0, 0], [0, 0, 0], [0, 0, 0]]
>>> [id(x) for x in matrix]
[32484168, 32484168, 32484168]
>>>

Three rows but only one object.

See the docs especially Note 2 about the s * n operation.

Fix:

>>> m2= [[0] * 3 for i in xrange(5)]
>>> [id(x) for x in m2]
[32498152, 32484808, 32498192, 32499952, 32499872]
>>>

Update: Here are some samples of code that gets an answer simply (i.e. without iter()):

>>> nrows = 2; ncols = 4
>>> zeroes = [[0 for j in xrange(ncols)] for i in xrange(nrows)]
>>> zeroes
[[0, 0, 0, 0], [0, 0, 0, 0]]
>>> ap = [[ncols * i + j for j in xrange(ncols)] for i in xrange(nrows)]
>>> ap
[[0, 1, 2, 3], [4, 5, 6, 7]]
>>>

2 of 3

Try running matrix[0][0] = 0 after that. Notice it now becomes:

[[0, 7, 8], [0, 7, 8], [0, 7, 8]]

So it's changing all three at the same time.

Read this: http://www.daniweb.com/software-development/python/threads/58916

That seems to explain it.