given the index of a row, l would like to remove that row.
l tried the following :
a.shape
Out[128]: (60, 3)
when l try to remove the row number 14 from my 3D array a as follow :
np.delete(a,14,axis=0)
a.shape
Out[130]: (60, 3)
l noticed that it doesn't make any chage. l supposed to get :
a.shape
Out[130]: (59, 3) # rather than (60,3)
What is wrong with my code ?
Assignment will solve it as apparently delete returns an array instead of working inplace:
a = np.delete(a,14,axis=0)
Related
In python, Suppose I have a 1D array C (c dimensions), and I want to construct a 4D matrix of dimension a x b x c x d, such that the array is duplicated along all other axes.
I.e. no matter what the dimension 1, 2 and 4 indexes are, array[i][j][k][l] = C[k]
Is there any numpy function to do that? Thanks!
For an array ar, you could use np.broadcast_to, to get that higher dim array as a view (hence virtually zero runtime and no memory overhead), like so -
np.broadcast_to(ar[None,None,:,None],(a,b,len(ar),d))
Sample run -
In [115]: ar = np.random.rand(10)
In [116]: a,b,d = 3,4,5
In [117]: np.broadcast_to(ar[None,None,:,None],(a,b,len(ar),d)).shape
Out[117]: (3, 4, 10, 5)
If you need output with its own memory space, append with .copy().
Leading newaxes(None) are optional. Hence, alternatively -
In [121]: np.broadcast_to(ar[:,None],(a,b,len(ar),d)).shape
Out[121]: (3, 4, 10, 5)
I want an array that looks like this,
array([array([[1, 1], [2, 2]]), array([3, 3])], dtype=object)
I can make an empty array and then assign elements one by one like this,
z = [np.array([[1,1],[2,2]]), np.array([3,3])]
x = np.empty(shape=2, dtype=object)
x[0], x[1] = z
I thought if this possible then so should be this: x = np.array(z, dtype=object), but that gets me the error: ValueError: could not broadcast input array from shape (2,2) into shape (2).
So is the way given above the only way to make a ragged numpy array? Or, is there a nice one line constructor/function we can can call to make the array x from above.
Say I have an image of 2x2 pixels named image_array, each pixel color is identified by a tuple of 3 entries (RGB), so the shape of image_array is 2x2x3.
I want to create an np.array c which has the shape 2x2x1 and which last coordinate is an empty set.
I tried this:
import numpy as np
image = (((1,2,3), (1,0,0)), ((1,1,1), (2,1,2)))
image_array = np.array(image)
c = np.empty(image_array.shape[:2], dtype=set)
c.fill(set())
c[0][1].add(124)
print(c)
I get:
[[{124} {124}]
[{124} {124}]]
And instead I would like the return:
[[{} {124}]
[{} {}]]
Any idea ?
The object array has to be filled with separate set() objects. That means creating them individually, as I do with a list comprehension:
In [279]: arr = np.array([set() for _ in range(4)]).reshape(2,2)
In [280]: arr
Out[280]:
array([[set(), set()],
[set(), set()]], dtype=object)
That construction should highlight that fact that this array is closely related to a list, or list of lists.
Now we can do a set operation on one of those elements:
In [281]: arr[0,1].add(124) # more idiomatic than arr[0][1]
In [282]: arr
Out[282]:
array([[set(), {124}],
[set(), set()]], dtype=object)
Note that we cannot operate on more than one set at a time. The object array offers few advantages compared to a list.
This is a 2d array; the sets don't form a dimension. Contrast that with
In [283]: image = (((1,2,3), (1,0,0)), ((1,1,1), (2,1,2)))
...: image_array = np.array(image)
...:
In [284]: image_array
Out[284]:
array([[[1, 2, 3],
[1, 0, 0]],
[[1, 1, 1],
[2, 1, 2]]])
While it started with tuples, it made a 3d array of integers.
Try this:
import numpy as np
x = np.empty((2, 2), dtype=np.object)
x[0, 0] = set(1, 2, 3)
print(x)
[[{1, 2, 3} None]
[None None]]
For non-number types in numpy you should use np.object.
whenever you do fill(set()), this will fill the array with exactly same set, as they refer to the same set. To fix this, just make a set if there isnt one everytime you need to add to the set
c = np.empty(image_array.shape[:2], dtype=set)
if not c[0][1]:
c[0,1] = set([124])
else:
c[0,1].add(124)
print (c)
# [[None {124}]
# [None None]]
Try changing your line c[0][1].add to this.
c[0][1] = 124
print(c)
This question already has answers here:
How do I add an extra column to a NumPy array?
(17 answers)
Closed 5 years ago.
l have the following vector
video_132.shape
Out[64]: (64, 3)
that l would to add to it a new 3D vector of three values
video_146[1][146][45]
such that
video_146[1][146][45].shape
Out[68]: (3,)
and
video_146[1][146][45]
Out[69]: array([217, 207, 198], dtype=uint8)
when l do the following
np.append(video_132,video_146[1][146][45])
l'm supposed to get
video_132.shape
Out[64]: (65, 3) # originally (64,3)
However l get :
Out[67]: (195,) # 64*3+3=195
It seems that it flattens the vector
How can l do the append by preserving the 3D structure ?
For visual simplicity let's rename video_132 --> a, and video_146[1][146][45] --> b. The particular values aren't important so let's say
In [82]: a = np.zeros((64, 3))
In [83]: b = np.ones((3,))
Then we can append b to a using:
In [84]: np.concatenate([a, b[None, :]]).shape
Out[84]: (65, 3)
Since np.concatenate returns a new array, reassign its return value to a to "append" b to a:
a = np.concatenate([a, b[None, :]])
Code for append:
def append(arr, values, axis=None):
arr = asanyarray(arr)
if axis is None:
if arr.ndim != 1:
arr = arr.ravel()
values = ravel(values)
axis = arr.ndim-1
return concatenate((arr, values), axis=axis)
Note how arr is raveled if no axis is provided
In [57]: np.append(np.ones((2,3)),2)
Out[57]: array([1., 1., 1., 1., 1., 1., 2.])
append is really aimed as simple cases like adding a scalar to a 1d array:
In [58]: np.append(np.arange(3),6)
Out[58]: array([0, 1, 2, 6])
Otherwise the behavior is hard to predict.
concatenate is the base operation (builtin) and takes a list, not just two. So we can collect many arrays (or lists) in one list and do one concatenate at the end of a loop. And since it doesn't tweak the dimensions before hand, it forces us to do that ourselves.
So to add a shape (3,) to a (64,3) we have transform that (3,) into (1,3). append requires the same dimension adjustment as concatenate if we specify the axis.
In [68]: np.append(arr,b[None,:], axis=0).shape
Out[68]: (65, 3)
In [69]: np.concatenate([arr,b[None,:]], axis=0).shape
Out[69]: (65, 3)
I need to read a text file that contains comma-delimited values into a 2D numpy array. The first 2 values on each line contain the index values for the numpy array and the third values contains the value to be stored in the array. As a catch, the index values are 1-based and need to be converted to the 0-based index values used by numpy. I've reviewed documentation and examples using genfromtxt and loadtxt but it's still not clear to me how to go about it. I've also tried the following code with no success:
a = np.arange(6).reshape(2,3)
for line in infile:
fields = line.split() #split fields inti list
rindex = int(fields[0]) - 1
cindex = int(fields[1]) - 1
a[rindex,cindex] = float(fields[2])
Here is an example of the input file:
1,1,10.1
1,2,11.2
1,3,12.3
2,3,13.4
2,2,14.5
2,3,15.6
And here is my desired output array. Ideally I'd like it to work on any array size without having to predefine the size of the array.
10.1 11.2 12.3
13.4 14.5 15.6
Here's one way you can do it. numpy.genfromtxt() is used to read the data into a structured array with three fields. The row and column indices are pulled out of the structured array and used to figure out the shape of the desired array, and to assign the values to the new array using numpy's "fancy" indexing:
In [46]: !cat test_data.csv
1,1,10.1
1,2,11.2
1,3,12.3
2,3,13.4
2,2,14.5
2,3,15.6
In [47]: data = np.genfromtxt('test_data.csv', dtype=None, delimiter=',', names=['i', 'j', 'value'])
In [48]: data
Out[48]:
array([(1, 1, 10.1), (1, 2, 11.2), (1, 3, 12.3), (2, 3, 13.4),
(2, 2, 14.5), (2, 3, 15.6)],
dtype=[('i', '<i8'), ('j', '<i8'), ('value', '<f8')])
In [49]: rows = data['i']
In [50]: cols = data['j']
In [51]: nrows = rows.max()
In [52]: ncols = cols.max()
In [53]: a = np.zeros((nrows, ncols))
In [54]: a[rows-1, cols-1] = data['value']
In [55]: a
Out[55]:
array([[ 10.1, 11.2, 12.3],
[ 0. , 14.5, 15.6]])