Is there way to initialize a numpy array of a shape and add to it? I will explain what I need with a list example. If I want to create a list of objects generated in a loop, I can do:
a = []
for i in range(5):
a.append(i)
I want to do something similar with a numpy array. I know about vstack, concatenate etc. However, it seems these require two numpy arrays as inputs. What I need is:
big_array # Initially empty. This is where I don't know what to specify
for i in range(5):
array i of shape = (2,4) created.
add to big_array
The big_array should have a shape (10,4). How to do this?
EDIT:
I want to add the following clarification. I am aware that I can define big_array = numpy.zeros((10,4)) and then fill it up. However, this requires specifying the size of big_array in advance. I know the size in this case, but what if I do not? When we use the .append function for extending the list in python, we don't need to know its final size in advance. I am wondering if something similar exists for creating a bigger array from smaller arrays, starting with an empty array.
numpy.fromiter() is what you are looking for:
big_array = numpy.fromiter(xrange(5), dtype="int")
It also works with generator expressions, e.g.:
big_array = numpy.fromiter( (i*(i+1)/2 for i in xrange(5)), dtype="int" )
If you know the length of the array in advance, you can specify it with an optional 'count' argument.
I'd suggest defining shape first. Then iterate over it to insert values.
big_array= np.zeros(shape = ( 6, 2 ))
for it in range(6):
big_array[it] = (it,it) # For example
>>>big_array
array([[ 0., 0.],
[ 1., 1.],
[ 2., 2.],
[ 3., 3.],
[ 4., 4.],
[ 5., 5.]])
You do want to avoid explicit loops as much as possible when doing array computing, as that reduces the speed gain from that form of computing. There are multiple ways to initialize a numpy array. If you want it filled with zeros, do as katrielalex said:
big_array = numpy.zeros((10,4))
EDIT: What sort of sequence is it you're making? You should check out the different numpy functions that create arrays, like numpy.linspace(start, stop, size) (equally spaced number), or numpy.arange(start, stop, inc). Where possible, these functions will make arrays substantially faster than doing the same work in explicit loops
Maybe something like this will fit your needs..
import numpy as np
N = 5
res = []
for i in range(N):
res.append(np.cumsum(np.ones(shape=(2,4))))
res = np.array(res).reshape((10, 4))
print(res)
Which produces the following output
[[ 1. 2. 3. 4.]
[ 5. 6. 7. 8.]
[ 1. 2. 3. 4.]
[ 5. 6. 7. 8.]
[ 1. 2. 3. 4.]
[ 5. 6. 7. 8.]
[ 1. 2. 3. 4.]
[ 5. 6. 7. 8.]
[ 1. 2. 3. 4.]
[ 5. 6. 7. 8.]]
Whenever you are in the following situation:
a = []
for i in range(5):
a.append(i)
and you want something similar in numpy, several previous answers have pointed out ways to do it, but as @katrielalex pointed out these methods are not efficient. The efficient way to do this is to build a long list and then reshape it the way you want after you have a long list. For example, let's say I am reading some lines from a file and each row has a list of numbers and I want to build a numpy array of shape (number of lines read, length of vector in each row). Here is how I would do it more efficiently:
long_list = []
counter = 0
with open('filename', 'r') as f:
for row in f:
row_list = row.split()
long_list.extend(row_list)
counter++
# now we have a long list and we are ready to reshape
result = np.array(long_list).reshape(counter, len(row_list)) # desired numpy array
The way I usually do that is by creating a regular list, then append my stuff into it, and finally transform the list to a numpy array as follows :
import numpy as np
big_array = [] # empty regular list
for i in range(5):
arr = i*np.ones((2,4)) # for instance
big_array.append(arr)
big_np_array = np.array(big_array) # transformed to a numpy array
of course your final object takes twice the space in the memory at the creation step, but appending on python list is very fast, and creation using np.array() also.
Introduced in numpy 1.8:
Return a new array of given shape and type, filled with fill_value.
Examples:
>>> import numpy as np
>>> np.full((2, 2), np.inf)
array([[ inf, inf],
[ inf, inf]])
>>> np.full((2, 2), 10)
array([[10, 10],
[10, 10]])
To initialize a numpy array with a specific matrix:
import numpy as np
mat = np.array([[1, 1, 0, 0, 0],
[0, 1, 0, 0, 1],
[1, 0, 0, 1, 1],
[0, 0, 0, 0, 0],
[1, 0, 1, 0, 1]])
print mat.shape
print mat
output:
(5, 5)
[[1 1 0 0 0]
[0 1 0 0 1]
[1 0 0 1 1]
[0 0 0 0 0]
[1 0 1 0 1]]
For your first array example use,
a = numpy.arange(5)
To initialize big_array, use
big_array = numpy.zeros((10,4))
This assumes you want to initialize with zeros, which is pretty typical, but there are many other ways to initialize an array in numpy.
Edit:
If you don't know the size of big_array in advance, it's generally best to first build a Python list using append, and when you have everything collected in the list, convert this list to a numpy array using numpy.array(mylist). The reason for this is that lists are meant to grow very efficiently and quickly, whereas numpy.concatenate would be very inefficient since numpy arrays don't change size easily. But once everything is collected in a list, and you know the final array size, a numpy array can be efficiently constructed.
Array analogue for the python's
a = []
for i in range(5):
a.append(i)
is:
import numpy as np
a = np.empty((0))
for i in range(5):
a = np.append(a, i)
I realize that this is a bit late, but I did not notice any of the other answers mentioning indexing into the empty array:
big_array = numpy.empty(10, 4)
for i in range(5):
array_i = numpy.random.random(2, 4)
big_array[2 * i:2 * (i + 1), :] = array_i
This way, you preallocate the entire result array with numpy.empty and fill in the rows as you go using indexed assignment.
It is perfectly safe to preallocate with empty instead of zeros in the example you gave since you are guaranteeing that the entire array will be filled with the chunks you generate.
Source: Stackoverflow.com