This question already has answers here:
How does assignment work with list slices?
(5 answers)
Closed 5 years ago.
I want to change ,multiple values in a list, for example, every multiple of 2. Using slicing.
my logic is:
list = [0] * 10
list[::2] = 1
However, I get an error message:
" must assign iterable to extended slice"
Can someone explain the error and also the correct logic to preform something like this? Thanks.
When you assign to a slice of a list, you need the assignment to be a list of the same length as the slice. For your example, assign a list of 5 ones:
l = [0] * 10
l[::2] = [1] *5
It isn't obvious why in this example, but if you think about it, you were doing:
l[3:6] = 2
Obviously that doesn't make sense. You are trying to assign an int to a list, which won't work. l[::2] is just another way to slice a list, so you must assign a list to it.
In the future, don't name your lists "list" because doing so overrides the builtin list() function.
my_list[::2] has 10//2 (=5) elements, so the right part of the assignment should have 10//2 elements as well:
>>> my_list = [0] * 10
>>> my_list[::2] = [1]*(10//2)
>>> my_list
[1, 0, 1, 0, 1, 0, 1, 0, 1, 0]
Or you could use numpy with broadcasting:
>>> import numpy as np
>>> a = np.zeros(10)
>>> a
array([ 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.])
>>> a[::2] = 1
>>> a
array([ 1., 0., 1., 0., 1., 0., 1., 0., 1., 0.])
Related
This question already has answers here:
Generate all binary strings of length n with k bits set
(14 answers)
Closed 15 days ago.
I have a question about how to generate all possible combinations of an array that satisfy the following condition:
with fixed length N
only M elements among N are 1, and the rest elements are 0.
For example, N = 4 and M = 2, we shall have
1: '0011'
2: '0101'
3: '0110'
4: '1001'
5: '1010'
6: '1100'
The basic idea is to pick up M elements from range(N) and replace the corresponding indices in np.zeros(N) with 1. However, I can only realize this idea by using several for-loops in python, which is inefficient. I'd like to ask whether there is any straightforward solution? Thanks in advance.
One way is to use itertools to get all possible combinations of the locations of ones and fill N-zero arrays with these ones.
import itertools
import numpy as np
N = 4
M = 2
combs = list(itertools.combinations(range(N), M))
result = [np.zeros(N) for _ in range(len(combs))]
for i, comb in enumerate(combs):
for j in comb:
result[i][j] = 1
print(result)
[array([1., 1., 0., 0.]), array([1., 0., 1., 0.]), array([1., 0., 0., 1.]), array([0., 1., 1., 0.]), array([0., 1., 0., 1.]), array([0., 0., 1., 1.])]
Generate a string with as many 0s and 1s as you need, then permutate it:
from itertools import permutations
n = 4
m = 2
item = []
for _ in range(m):
item.append("1")
for _ in range(n-m):
item.append("0")
perm = permutations(item)
for x in perm:
my_str = ""
for c in x:
my_str += c
print(my_str)
I have the following question. Is there somekind of method with numpy or scipy , which I can use to get an given unsorted array like this
a = np.array([0,0,1,1,4,4,4,4,5,1891,7]) #could be any number here
to something where the numbers are interpolated/mapped , there is no gap between the values and they are in the same order like before?:
[0,0,1,1,2,2,2,2,3,5,4]
EDIT
Is it furthermore possible to swap/shuffle the numbers after the mapping, so that
[0,0,1,1,2,2,2,2,3,5,4]
become something like:
[0,0,3,3,5,5,5,5,4,1,2]
Edit: I'm not sure what the etiquette is here (should this be a separate answer?), but this is actually directly obtainable from np.unique.
>>> u, indices = np.unique(a, return_inverse=True)
>>> indices
array([0, 0, 1, 1, 2, 2, 2, 2, 3, 5, 4])
Original answer: This isn't too hard to do in plain python by building a dictionary of what index each value of the array would map to:
x = np.sort(np.unique(a))
index_dict = {j: i for i, j in enumerate(x)}
[index_dict[i] for i in a]
Seems you need to rank (dense) your array, in which case use scipy.stats.rankdata:
from scipy.stats import rankdata
rankdata(a, 'dense')-1
# array([ 0., 0., 1., 1., 2., 2., 2., 2., 3., 5., 4.])
I have a 2D array, and it has some duplicate columns. I would like to be able to see which unique columns there are, and where the duplicates are.
My own array is too large to put here, but here is an example:
a = np.array([[ 1., 0., 0., 0., 0.],[ 2., 0., 4., 3., 0.],])
This has the unique column vectors [1.,2.], [0.,0.], [0.,4.] and [0.,3.]. There is one duplicate: [0.,0.] appears twice.
Now I found a way to get the unique vectors and their indices here but it is not clear to me how I would get the occurences of duplicates as well. I have tried several naive ways (with np.where and list comps) but those are all very very slow. Surely there has to be a numpythonic way?
In matlab it's just the unique function but np.unique flattens arrays.
Here's a vectorized approach to give us a list of arrays as output -
ids = np.ravel_multi_index(a.astype(int),a.max(1).astype(int)+1)
sidx = ids.argsort()
sorted_ids = ids[sidx]
out = np.split(sidx,np.nonzero(sorted_ids[1:] > sorted_ids[:-1])[0]+1)
Sample run -
In [62]: a
Out[62]:
array([[ 1., 0., 0., 0., 0.],
[ 2., 0., 4., 3., 0.]])
In [63]: out
Out[63]: [array([1, 4]), array([3]), array([2]), array([0])]
The numpy_indexed package (disclaimer: I am its author) contains efficient functionality for computing these kind of things:
import numpy_indexed as npi
unique_columns = npi.unique(a, axis=1)
non_unique_column_idx = npi.multiplicity(a, axis=1) > 1
Or alternatively:
unique_columns, column_count = npi.count(a, axis=1)
duplicate_columns = unique_columns[:, column_count > 1]
For small arrays:
from collections import defaultdict
indices = defaultdict(list)
for index, column in enumerate(a.transpose()):
indices[tuple(column)].append(index)
unique = [kk for kk, vv in indices.items() if len(vv) == 1]
non_unique = {kk:vv for kk, vv in indices.items() if len(vv) != 1}
I am trying to fill an empty(not np.empty!) array with values using append but I am gettin error:
My code is as follows:
import numpy as np
result=np.asarray([np.asarray([]),np.asarray([])])
result[0]=np.append([result[0]],[1,2])
And I am getting:
ValueError: could not broadcast input array from shape (2) into shape (0)
I might understand the question incorrectly, but if you want to declare an array of a certain shape but with nothing inside, the following might be helpful:
Initialise empty array:
>>> a = np.zeros((0,3)) #or np.empty((0,3)) or np.array([]).reshape(0,3)
>>> a
array([], shape=(0, 3), dtype=float64)
Now you can use this array to append rows of similar shape to it. Remember that a numpy array is immutable, so a new array is created for each iteration:
>>> for i in range(3):
... a = np.vstack([a, [i,i,i]])
...
>>> a
array([[ 0., 0., 0.],
[ 1., 1., 1.],
[ 2., 2., 2.]])
np.vstack and np.hstack is the most common method for combining numpy arrays, but coming from Matlab I prefer np.r_ and np.c_:
Concatenate 1d:
>>> a = np.zeros(0)
>>> for i in range(3):
... a = np.r_[a, [i, i, i]]
...
>>> a
array([ 0., 0., 0., 1., 1., 1., 2., 2., 2.])
Concatenate rows:
>>> a = np.zeros((0,3))
>>> for i in range(3):
... a = np.r_[a, [[i,i,i]]]
...
>>> a
array([[ 0., 0., 0.],
[ 1., 1., 1.],
[ 2., 2., 2.]])
Concatenate columns:
>>> a = np.zeros((3,0))
>>> for i in range(3):
... a = np.c_[a, [[i],[i],[i]]]
...
>>> a
array([[ 0., 1., 2.],
[ 0., 1., 2.],
[ 0., 1., 2.]])
numpy.append is pretty different from list.append in python. I know that's thrown off a few programers new to numpy. numpy.append is more like concatenate, it makes a new array and fills it with the values from the old array and the new value(s) to be appended. For example:
import numpy
old = numpy.array([1, 2, 3, 4])
new = numpy.append(old, 5)
print old
# [1, 2, 3, 4]
print new
# [1, 2, 3, 4, 5]
new = numpy.append(new, [6, 7])
print new
# [1, 2, 3, 4, 5, 6, 7]
I think you might be able to achieve your goal by doing something like:
result = numpy.zeros((10,))
result[0:2] = [1, 2]
# Or
result = numpy.zeros((10, 2))
result[0, :] = [1, 2]
Update:
If you need to create a numpy array using loop, and you don't know ahead of time what the final size of the array will be, you can do something like:
import numpy as np
a = np.array([0., 1.])
b = np.array([2., 3.])
temp = []
while True:
rnd = random.randint(0, 100)
if rnd > 50:
temp.append(a)
else:
temp.append(b)
if rnd == 0:
break
result = np.array(temp)
In my example result will be an (N, 2) array, where N is the number of times the loop ran, but obviously you can adjust it to your needs.
new update
The error you're seeing has nothing to do with types, it has to do with the shape of the numpy arrays you're trying to concatenate. If you do np.append(a, b) the shapes of a and b need to match. If you append an (2, n) and (n,) you'll get a (3, n) array. Your code is trying to append a (1, 0) to a (2,). Those shapes don't match so you get an error.
This error arise from the fact that you are trying to define an object of shape (0,) as an object of shape (2,). If you append what you want without forcing it to be equal to result[0] there is no any issue:
b = np.append([result[0]], [1,2])
But when you define result[0] = b you are equating objects of different shapes, and you can not do this. What are you trying to do?
Here's the result of running your code in Ipython. Note that result is a (2,0) array, 2 rows, 0 columns, 0 elements. The append produces a (2,) array. result[0] is (0,) array. Your error message has to do with trying to assign that 2 item array into a size 0 slot. Since result is dtype=float64, only scalars can be assigned to its elements.
In [65]: result=np.asarray([np.asarray([]),np.asarray([])])
In [66]: result
Out[66]: array([], shape=(2, 0), dtype=float64)
In [67]: result[0]
Out[67]: array([], dtype=float64)
In [68]: np.append(result[0],[1,2])
Out[68]: array([ 1., 2.])
np.array is not a Python list. All elements of an array are the same type (as specified by the dtype). Notice also that result is not an array of arrays.
Result could also have been built as
ll = [[],[]]
result = np.array(ll)
while
ll[0] = [1,2]
# ll = [[1,2],[]]
the same is not true for result.
np.zeros((2,0)) also produces your result.
Actually there's another quirk to result.
result[0] = 1
does not change the values of result. It accepts the assignment, but since it has 0 columns, there is no place to put the 1. This assignment would work in result was created as np.zeros((2,1)). But that still can't accept a list.
But if result has 2 columns, then you can assign a 2 element list to one of its rows.
result = np.zeros((2,2))
result[0] # == [0,0]
result[0] = [1,2]
What exactly do you want result to look like after the append operation?
numpy.append always copies the array before appending the new values. Your code is equivalent to the following:
import numpy as np
result = np.zeros((2,0))
new_result = np.append([result[0]],[1,2])
result[0] = new_result # ERROR: has shape (2,0), new_result has shape (2,)
Perhaps you mean to do this?
import numpy as np
result = np.zeros((2,0))
result = np.append([result[0]],[1,2])
SO thread 'Multiply two arrays element wise, where one of the arrays has arrays as elements' has an example of constructing an array from arrays. If the subarrays are the same size, numpy makes a 2d array. But if they differ in length, it makes an array with dtype=object, and the subarrays retain their identity.
Following that, you could do something like this:
In [5]: result=np.array([np.zeros((1)),np.zeros((2))])
In [6]: result
Out[6]: array([array([ 0.]), array([ 0., 0.])], dtype=object)
In [7]: np.append([result[0]],[1,2])
Out[7]: array([ 0., 1., 2.])
In [8]: result[0]
Out[8]: array([ 0.])
In [9]: result[0]=np.append([result[0]],[1,2])
In [10]: result
Out[10]: array([array([ 0., 1., 2.]), array([ 0., 0.])], dtype=object)
However, I don't offhand see what advantages this has over a pure Python list or lists. It does not work like a 2d array. For example I have to use result[0][1], not result[0,1]. If the subarrays are all the same length, I have to use np.array(result.tolist()) to produce a 2d array.
So I have a string of RGBA image data, each pixel is a byte long. I know the image's x and y resolution too. Now I want to edit the string in a way which would cause the image to be flipped or reversed vertically, which means have the first "row" of pixels become the last row and the opposite, and like this for all other "rows". Is there a fast way to do it?
To do what you want to the letter this is one way to proceed:
>>> img = 'ABCDEFGHIJKL'
>>> x, y = 4, 3
>>> def chunks(l, n):
... for i in xrange(0, len(l), n):
... yield l[i:i+n]
...
>>> [row for row in chunks(img, x)]
['ABCD', 'EFGH', 'IJKL']
>>> ''.join(reversed([row for row in chunks(img, x)]))
'IJKLEFGHABCD'
HOWEVER, unless you have very small images, you would be better off passing through numpy, as this is at the very minimum an order of magnitude faster than Cpython datatypes. You should look at at the flipup function. Example:
>>> A
array([[ 1., 0., 0.],
[ 0., 2., 0.],
[ 0., 0., 3.]])
>>> np.flipud(A)
array([[ 0., 0., 3.],
[ 0., 2., 0.],
[ 1., 0., 0.]])
EDIT: thought to add a complete example in case you have never worked with NumPy before. Of course the conversion is worth only for images that are not 2x2, as instantiating the array has an overhead....
>>> import numpy as np
>>> img = [0x00, 0x01, 0x02, 0x03]
>>> img
[0, 1, 2, 3]
>>> x = y = 2
>>> aimg = np.array(img).reshape(x, y)
>>> aimg
array([[0, 1],
[2, 3]])
>>> np.flipud(aimg)
array([[2, 3],
[0, 1]])
say you have the image in array img, then do
img.reverse();
#also need to flip each row
for row in img:
row.reverse();