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I want to convert a list of tuples into a numpy array. For example:

items = [(1, 2), (3, 4)]

using np.asarray(items) I get:

array([[1, 2],
 [3, 4]])

but if I try to append the items individually:

new_array = np.empty(0)
for item in items:
 new_array = np.append(new_array, item)

the new_array loses the original shape and becomes:

array([1., 2., 3., 4.])

I can get it to the shape I wanted using new_array.reshape(2, 2):

array([[1., 2.],
 [3., 4.]])

but how would I get that shape without reshaping?

Firstly you need to provide a correct shape to the array so that numpy could understand how to interpret the values provided to the append method.

Then, to prevent automatic flattening, specify the axis you wish to append on.

This code does what you intended to do:

import numpy as np

items = [(1,2),(3,4)]

new_array = np.ndarray((0,2))
for item in items:
 new_array = np.append(new_array, [item], axis=0)

print(new_array) # [[1. 2.]
 # [3. 4.]]

If you have a list of tuples, and you've decided you hate the standard array constructors (np.array, np.asarray, etc, which, as @JohnZwinck pointed out are probably the best answer) for some reason, the most efficient approach would be to preallocate the entire array and then assign to it:

items = [(1, 2), (3, 4)]
arr = np.empty((len(items), len(items[0])))

arr[...] = items

Even if what you want is to grow an array over time, row-by-row, it has been shown through detailed timings that you're usually better off just allocating a whole new array and then copying over the old values.

So given the above arr, by this approach the most efficient way to append a row would be:

newitem = (5, 6)
oldarr = arr
arr = np.empty((oldarr.shape[0] + 1, *oldarr.shape[1:]))

arr[:-1,:] = oldarr
arr[-1,:] = newitem