CODEWITHSUNDEEP
pythonarraysnumpy
user1220022
user1220022

Reputation: 12075

Initialise numpy array of unknown length

I want to be able to 'build' a numpy array on the fly, I do not know the size of this array in advance.

For example I want to do something like this:

a= np.array()
for x in y:
     a.append(x)

Which would result in a containing all the elements of x, obviously this is a trivial answer. I am just curious whether this is possible?

Upvotes: 77

Views: 232781

Answers (7)

Parmar Aayush
Parmar Aayush

Reputation: 11

list1 = []
size = 1
option = "Y"
for x in range(size):
    ele = input("Enter Element For List One : ")
    list1.append(ele)
while(option == "Y"):
    option = input("\n***Add More Element Press Y ***: ")
    if(option=="Y"):
        size = size + 1
        for x in range(size):
            ele = input("Enter Element For List Element : ")
            list1.append(ele)
            size = 1
    else:
        break;
print(list1)

Take:

list1 = []    # Store Array Element 
size = 1      # Rune at One Time 
option = "Y"  # Take User Choice

Implementation:

  • For Loop at a range of size variable which runs one time because size = 1
  • use While loop
    • take input from users if they want to add press "Y" else "N"
    • in the while loop use the if else statement if the option is "Y" then increase size with which helps run 2 time array because size = size + 1 run another for loop at a range of size
    • else break

Upvotes: -1

Shubham Agrawal
Shubham Agrawal

Reputation: 182

I wrote a small utility function. (most answers above are good. I feel this looks nicer)

def np_unknown_cat(acc, arr):
  arrE = np.expand_dims(arr, axis=0)
  if acc is None:
    return arrE
  else:
    return np.concatenate((acc, arrE))

You can use the above function as the following:

acc = None  # accumulator
arr1 = np.ones((3,4))
acc = np_unknown_cat(acc, arr1)
arr2 = np.ones((3,4))
acc = np_unknown_cat(acc, arr2)

Upvotes: 0

Mr_and_Mrs_D
Mr_and_Mrs_D

Reputation: 34016

Since y is an iterable I really do not see why the calls to append:

a = np.array(list(y))

will do and it's much faster:

import timeit

print timeit.timeit('list(s)', 's=set(x for x in xrange(1000))')
# 23.952975494633154

print timeit.timeit("""li=[]
for x in s: li.append(x)""", 's=set(x for x in xrange(1000))')
# 189.3826994248866

Upvotes: 7

chiefenne
chiefenne

Reputation: 605

a = np.empty(0)
for x in y:
    a = np.append(a, x)

Upvotes: 1

BenDundee
BenDundee

Reputation: 4521

For posterity, I think this is quicker:

a = np.array([np.array(list()) for _ in y])

You might even be able to pass in a generator (i.e. [] -> ()), in which case the inner list is never fully stored in memory.

Responding to comment below:

>>> import numpy as np
>>> y = range(10)
>>> a = np.array([np.array(list) for _ in y])
>>> a
array([array(<type 'list'>, dtype=object),
       array(<type 'list'>, dtype=object),
       array(<type 'list'>, dtype=object),
       array(<type 'list'>, dtype=object),
       array(<type 'list'>, dtype=object),
       array(<type 'list'>, dtype=object),
       array(<type 'list'>, dtype=object),
       array(<type 'list'>, dtype=object),
       array(<type 'list'>, dtype=object),
       array(<type 'list'>, dtype=object)], dtype=object)

Upvotes: 1

Fred Foo
Fred Foo

Reputation: 363547

Build a Python list and convert that to a Numpy array. That takes amortized O(1) time per append + O(n) for the conversion to array, for a total of O(n).

    a = []
    for x in y:
        a.append(x)
    a = np.array(a)

Upvotes: 130

alexisdm
alexisdm

Reputation: 29886

You can do this:

a = np.array([])
for x in y:
    a = np.append(a, x)

Upvotes: 19

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