I have an array of numpy arrays:
a = [[1, 2, 3, 4], [1, 2, 3, 5], [2, 5, 4, 3], [5, 2, 3, 1]]
I need to find and remove a particular list from a:
rem = [1,2,3,5]
numpy.delete(a,rem) does not return the correct results. I need to be able to return:
[[1, 2, 3, 4], [2, 5, 4, 3], [5, 2, 3, 1]]
is this possible with numpy?
A list comprehension can achieve this.
rem = [1,2,3,5]
a = [[1, 2, 3, 4], [1, 2, 3, 5], [2, 5, 4, 3], [5, 2, 3, 1]]
a = [x for x in a if x != rem]
outputs
[[1, 2, 3, 4], [2, 5, 4, 3], [5, 2, 3, 1]]
Numpy arrays do not support random deletion by element. Similar to strings in Python, you need to generate a new array to delete a single or multiple sub elements.
Given:
>>> a
array([[1, 2, 3, 4],
[1, 2, 3, 5],
[2, 5, 4, 3],
[5, 2, 3, 1]])
>>> rem
array([1, 2, 3, 5])
You can get each matching sub array and create a new array from that:
>>> a=np.array([sa for sa in a if not np.all(sa==rem)])
>>> a
array([[1, 2, 3, 4],
[2, 5, 4, 3],
[5, 2, 3, 1]])
To use np.delete, you would use an index and not a match, so:
>>> a
array([[1, 2, 3, 4],
[1, 2, 3, 5],
[2, 5, 4, 3],
[5, 2, 3, 1]])
>>> np.delete(a, 1, 0) # delete element 1, axis 0
array([[1, 2, 3, 4],
[2, 5, 4, 3],
[5, 2, 3, 1]])
But you can't loop over the array and delete elements...
You can pass multiple elements to np.delete however and you just need to match sub elements:
>>> a
array([[1, 2, 3, 5],
[1, 2, 3, 5],
[2, 5, 4, 3],
[5, 2, 3, 1]])
>>> np.delete(a, [i for i, sa in enumerate(a) if np.all(sa==rem)], 0)
array([[2, 5, 4, 3],
[5, 2, 3, 1]])
And given that same a, you can have an all numpy solution by using np.where:
>>> np.delete(a, np.where((a == rem).all(axis=1)), 0)
array([[2, 5, 4, 3],
[5, 2, 3, 1]])
Did you try list remove?
In [84]: a = [[1, 2, 3, 4], [1, 2, 3, 5], [2, 5, 4, 3], [5, 2, 3, 1]]
In [85]: a
Out[85]: [[1, 2, 3, 4], [1, 2, 3, 5], [2, 5, 4, 3], [5, 2, 3, 1]]
In [86]: rem = [1,2,3,5]
In [87]: a.remove(rem)
In [88]: a
Out[88]: [[1, 2, 3, 4], [2, 5, 4, 3], [5, 2, 3, 1]]
remove matches on value.
np.delete works with an index, not value. Also it returns a copy; it does not act in place. And the result is an array, not a nested list (np.delete converts the input to an array before operating on it).
In [92]: a = [[1, 2, 3, 4], [1, 2, 3, 5], [2, 5, 4, 3], [5, 2, 3, 1]]
In [93]: a1=np.delete(a,1, axis=0)
In [94]: a1
Out[94]:
array([[1, 2, 3, 4],
[2, 5, 4, 3],
[5, 2, 3, 1]])
This is more like list pop:
In [96]: a = [[1, 2, 3, 4], [1, 2, 3, 5], [2, 5, 4, 3], [5, 2, 3, 1]]
In [97]: a.pop(1)
Out[97]: [1, 2, 3, 5]
In [98]: a
Out[98]: [[1, 2, 3, 4], [2, 5, 4, 3], [5, 2, 3, 1]]
To delete by value you need first find the index of the desired row. With integer arrays that's not too hard. With floats it is trickier.
=========
But you don't need to use delete to do this in numpy; boolean indexing works:
In [119]: a = [[1, 2, 3, 4], [1, 2, 3, 5], [2, 5, 4, 3], [5, 2, 3, 1]]
In [120]: A = np.array(a) # got to work with array, not list
In [121]: rem=np.array([1,2,3,5])
Simple comparison; rem is broadcasted to match rows
In [122]: A==rem
Out[122]:
array([[ True, True, True, False],
[ True, True, True, True],
[False, False, False, False],
[False, True, True, False]], dtype=bool)
find the row where all elements match - this is the one we want to remove
In [123]: (A==rem).all(axis=1)
Out[123]: array([False, True, False, False], dtype=bool)
Just not it, and use it to index A:
In [124]: A[~(A==rem).all(axis=1),:]
Out[124]:
array([[1, 2, 3, 4],
[2, 5, 4, 3],
[5, 2, 3, 1]])
(the original A is not changed).
np.where can be used to convert the boolean (or its inverse) to indicies. Sometimes that's handy, but usually it isn't required.
Related
Input: A 2D numpy array
Output: An array of indices that will sort the array row by row (or column by column)
E.g.: Say the function is get_sorted_indices(array, axis=0)
a = np.array([[1,2,3,4,5]
,[2,3,4,5,6]
,[1,2,3,4,5]
,[2,3,4,6,6]
,[2,3,4,5,6]])
ind = get_sorted_indices(a, axis=0)
Then we will get
>>> ind
[0, 2, 1, 4, 3]
>>> a[ind] # should be equals to a.sort(axis = 0)
array([[1, 2, 3, 4, 5],
[1, 2, 3, 4, 5],
[2, 3, 4, 5, 6],
[2, 3, 4, 5, 6],
[2, 3, 4, 6, 6]])
>>> a.sort(axis=0)
array([[1, 2, 3, 4, 5],
[1, 2, 3, 4, 5],
[2, 3, 4, 5, 6],
[2, 3, 4, 5, 6],
[2, 3, 4, 6, 6]])
I've looked at argsort but I don't understand its output and reading the documentation doesn't help:
>>> a.argsort()
array([[0, 1, 2, 3, 4],
[0, 1, 2, 3, 4],
[0, 1, 2, 3, 4],
[0, 1, 2, 3, 4],
[0, 1, 2, 3, 4]])
>>> a.argsort(axis=0)
array([[0, 0, 0, 0, 0],
[2, 2, 2, 2, 2],
[1, 1, 1, 1, 1],
[3, 3, 3, 4, 3],
[4, 4, 4, 3, 4]])
I can do this manually but I think I'm misunderstanding argsort or I'm missing something from numpy.
Is there a standard way to do this or I have no choice but to do this manually?
I'm a MatLab user who recently converted to python. I am running a for loop that cuts a longer signal into individual trials, normalizes them to 100% trial and then would like to have the trials listed horizontally in a single variable. My code is
RHipFE=np.empty([101, 1])
newlength = 101
for i in range(0,len(R0X)-1,2):
iHipFE=redataf.RHipFE[R0X[i]:R0X[i+1]]
x=np.arange(0,len(iHipFE),1)
new_x = np.linspace(x.min(), x.max(), newlength)
iHipFEn = interpolate.interp1d(x, iHipFE)(new_x)
RHipFE=np.concatenate((RHipFE,iHipFEn),axis=1)
When I run this, I get the error "ValueError: all the input arrays must have same number of dimensions". Which I assume is because RHipFE is (101,1) while iHipFEn is (101,). Is the best solution to make iHipFEn (101,1)? If so, how does one do this in the above for loop?
Generally it's faster to collect arrays in a list, and use some form of concatenate once. List append is faster than concatenate:
In [51]: alist = []
In [52]: for i in range(3):
...: alist.append(np.arange(i,i+5))
...:
In [53]: alist
Out[53]: [array([0, 1, 2, 3, 4]), array([1, 2, 3, 4, 5]), array([2, 3, 4, 5, 6])]
Various ways of joining
In [54]: np.vstack(alist)
Out[54]:
array([[0, 1, 2, 3, 4],
[1, 2, 3, 4, 5],
[2, 3, 4, 5, 6]])
In [55]: np.column_stack(alist)
Out[55]:
array([[0, 1, 2],
[1, 2, 3],
[2, 3, 4],
[3, 4, 5],
[4, 5, 6]])
In [56]: np.stack(alist, axis=1)
Out[56]:
array([[0, 1, 2],
[1, 2, 3],
[2, 3, 4],
[3, 4, 5],
[4, 5, 6]])
In [57]: np.array(alist)
Out[57]:
array([[0, 1, 2, 3, 4],
[1, 2, 3, 4, 5],
[2, 3, 4, 5, 6]])
Internally, vstack, column_stack, stack expand the dimension of the components, and concatenate on the appropriate axis:
In [58]: np.concatenate([l[:,None] for l in alist],axis=1)
Out[58]:
array([[0, 1, 2],
[1, 2, 3],
[2, 3, 4],
[3, 4, 5],
[4, 5, 6]])
How do I use numpy / python array routines to do this ?
E.g. If I have array [ [1,2,3,4,]] , the output should be
[[1,1,2,2,],
[1,1,2,2,],
[3,3,4,4,],
[3,3,4,4]]
Thus, the output is array of double the row and column dimensions. And each element from original array is repeated three times.
What I have so far is this
def operation(mat,step=2):
result = np.array(mat,copy=True)
result[::2,::2] = mat
return result
This gives me array
[[ 98.+0.j 0.+0.j 40.+0.j 0.+0.j]
[ 0.+0.j 0.+0.j 0.+0.j 0.+0.j]
[ 29.+0.j 0.+0.j 54.+0.j 0.+0.j]
[ 0.+0.j 0.+0.j 0.+0.j 0.+0.j]]
for the input
[[98 40]
[29 54]]
The array will always be of even dimensions.
Use np.repeat():
In [9]: A = np.array([[1, 2, 3, 4]])
In [10]: np.repeat(np.repeat(A, 2).reshape(2, 4), 2, 0)
Out[10]:
array([[1, 1, 2, 2],
[1, 1, 2, 2],
[3, 3, 4, 4],
[3, 3, 4, 4]])
Explanation:
First off you can repeat the arrya items:
In [30]: np.repeat(A, 3)
Out[30]: array([1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4])
then all you need is reshaping the result (based on your expected result this can be different):
In [32]: np.repeat(A, 3).reshape(2, 3*2)
array([[1, 1, 1, 2, 2, 2],
[3, 3, 3, 4, 4, 4]])
And now you should repeat the result along the the first axis:
In [34]: np.repeat(np.repeat(A, 3).reshape(2, 3*2), 3, 0)
Out[34]:
array([[1, 1, 1, 2, 2, 2],
[1, 1, 1, 2, 2, 2],
[1, 1, 1, 2, 2, 2],
[3, 3, 3, 4, 4, 4],
[3, 3, 3, 4, 4, 4],
[3, 3, 3, 4, 4, 4]])
Another approach could be with np.kron -
np.kron(a.reshape(-1,2),np.ones((2,2),dtype=int))
Basically, we reshape input array into a 2D array keeping the second axis of length=2. Then np.kron essentially replicates the elements along both rows and columns for a length of 2 each with that array : np.ones((2,2),dtype=int).
Sample run -
In [45]: a
Out[45]: array([7, 5, 4, 2, 8, 6])
In [46]: np.kron(a.reshape(-1,2),np.ones((2,2),dtype=int))
Out[46]:
array([[7, 7, 5, 5],
[7, 7, 5, 5],
[4, 4, 2, 2],
[4, 4, 2, 2],
[8, 8, 6, 6],
[8, 8, 6, 6]])
If you would like to have 4 rows, use a.reshape(2,-1) instead.
The better solution is to use numpy but you could use iteration also:
a = [[1, 2, 3, 4]]
v = iter(a[0])
b = []
for i in v:
n = next(v)
[b.append([i for k in range(2)] + [n for k in range(2)]) for j in range(2)]
print b
>>> [[1, 1, 2, 2], [1, 1, 2, 2], [3, 3, 4, 4], [3, 3, 4, 4]]
I have a large 2d array of vectors. I want to split this array into several arrays according to one of the vectors' elements or dimensions. I would like to receive one such small array if the values along this column are consecutively identical. For example considering the third dimension or column:
orig = np.array([[1, 2, 3],
[3, 4, 3],
[5, 6, 4],
[7, 8, 4],
[9, 0, 4],
[8, 7, 3],
[6, 5, 3]])
I want to turn into three arrays consisting of rows 1,2 and 3,4,5 and 6,7:
>>> a
array([[1, 2, 3],
[3, 4, 3]])
>>> b
array([[5, 6, 4],
[7, 8, 4],
[9, 0, 4]])
>>> c
array([[8, 7, 3],
[6, 5, 3]])
I'm new to python and numpy. Any help would be greatly appreciated.
Regards
Mat
Edit: I reformatted the arrays to clarify the problem
Using np.split:
>>> a, b, c = np.split(orig, np.where(orig[:-1, 2] != orig[1:, 2])[0]+1)
>>> a
array([[1, 2, 3],
[1, 2, 3]])
>>> b
array([[1, 2, 4],
[1, 2, 4],
[1, 2, 4]])
>>> c
array([[1, 2, 3],
[1, 2, 3]])
Nothing fancy here, but this good old-fashioned loop should do the trick
import numpy as np
a = np.array([[1, 2, 3],
[1, 2, 3],
[1, 2, 4],
[1, 2, 4],
[1, 2, 4],
[1, 2, 3],
[1, 2, 3]])
groups = []
rows = a[0]
prev = a[0][-1] # here i assume that the grouping is based on the last column, change the index accordingly if that is not the case.
for row in a[1:]:
if row[-1] == prev:
rows = np.vstack((rows, row))
else:
groups.append(rows)
rows = [row]
prev = row[-1]
groups.append(rows)
print groups
## [array([[1, 2, 3],
## [1, 2, 3]]),
## array([[1, 2, 4],
## [1, 2, 4],
## [1, 2, 4]]),
## array([[1, 2, 3],
## [1, 2, 3]])]
if a looks like this:
array([[1, 1, 2, 3],
[2, 1, 2, 3],
[3, 1, 2, 4],
[4, 1, 2, 4],
[5, 1, 2, 4],
[6, 1, 2, 3],
[7, 1, 2, 3]])
than this
col = a[:, -1]
indices = np.where(col[:-1] != col[1:])[0] + 1
indices = np.concatenate(([0], indices, [len(a)]))
res = [a[start:end] for start, end in zip(indices[:-1], indices[1:])]
print(res)
results in:
[array([[1, 2, 3],
[1, 2, 3]]), array([[1, 2, 4],
[1, 2, 4],
[1, 2, 4]]), array([[1, 2, 3],
[1, 2, 3]])]
Update: np.split() is much nicer. No need to add first and last index:
col = a[:, -1]
indices = np.where(col[:-1] != col[1:])[0] + 1
res = np.split(a, indices)
Given the following example
d = array([[1, 2, 3],
[1, 2, 3],
[1, 3, 3],
[4, 4, 4],
[5, 5, 5]
])
To get the sub-array containing 1 in the first column:
d[ d[:,0] == 1 ]
array([[1, 2, 3],
[1, 2, 3],
[1, 3, 3]])
How to get (without loops) the sub-array containing 1 and 5? Shouldn't be something like
d[ d[:,0] == [1,5] ] # ---> array([1, 2, 3])
which does not work?
Method #1: use bitwise or | to combine the conditions:
>>> d
array([[1, 2, 3],
[1, 2, 3],
[1, 3, 3],
[4, 4, 4],
[5, 5, 5]])
>>> (d[:,0] == 1) | (d[:,0] == 5)
array([ True, True, True, False, True], dtype=bool)
>>> d[(d[:,0] == 1) | (d[:,0] == 5)]
array([[1, 2, 3],
[1, 2, 3],
[1, 3, 3],
[5, 5, 5]])
Method #2: use np.in1d, which is probably easier if there are a lot of values:
>>> np.in1d(d[:,0], [1, 5])
array([ True, True, True, False, True], dtype=bool)
>>> d[np.in1d(d[:,0], [1, 5])]
array([[1, 2, 3],
[1, 2, 3],
[1, 3, 3],
[5, 5, 5]])