>>> arr = np.array([[1,2,3,4], [5,6,7,8], [9,10,11,12]])
>>> arr
array([[ 1, 2, 3, 4],
[ 5, 6, 7, 8],
[ 9, 10, 11, 12]])
I am deleting the 3rd column as
>>> np.hstack(((np.delete(arr, np.s_[2:], 1)),(np.delete(arr, np.s_[:3],1))))
array([[ 1, 2, 4],
[ 5, 6, 8],
[ 9, 10, 12]])
Are there any better way ?
Please consider this to be a novice question.
If you ever want to delete more than one columns, you just pass indices of columns you want deleted as a list, like this:
>>> a = np.arange(12).reshape(3,4)
>>> a
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
>>> np.delete(a, [1,3], axis=1)
array([[ 0, 2],
[ 4, 6],
[ 8, 10]])
>>> import numpy as np
>>> arr = np.array([[1,2,3,4], [5,6,7,8], [9,10,11,12]])
>>> np.delete(arr, 2, axis=1)
array([[ 1, 2, 4],
[ 5, 6, 8],
[ 9, 10, 12]])
Something like this:
In [7]: x = range(16)
In [8]: x = np.reshape(x, (4, 4))
In [9]: x
Out[9]:
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15]])
In [10]: np.delete(x, 1, 1)
Out[10]:
array([[ 0, 2, 3],
[ 4, 6, 7],
[ 8, 10, 11],
[12, 14, 15]])
Related
I have a pandas data frame that looks like this:
>>> df = pd.DataFrame({'a': list(range(10))})
>>> df['a'] = df.a.apply(lambda x: x*np.array([1,2,3]))
>>>df.head()
a
0 [0, 0, 0]
1 [1, 2, 3]
2 [2, 4, 6]
3 [3, 6, 9]
4 [4, 8, 12]
I would like to get column a from the df as a ndarray. But when I do that I get an array of arrays
>>> df.a.values
array([array([0, 0, 0]), array([1, 2, 3]), array([2, 4, 6]),
array([3, 6, 9]), array([ 4, 8, 12]), array([ 5, 10, 15]),
array([ 6, 12, 18]), array([ 7, 14, 21]), array([ 8, 16, 24]),
array([ 9, 18, 27])], dtype=object)
How can I get the returnd output to be
array([[ 0, 0, 0],
[ 1, 2, 3],
[ 2, 4, 6],
[ 3, 6, 9],
[ 4, 8, 12],
# ...
])
Using pandas,
df.a.apply(pd.Series).values
Using numpy,
np.vstack(df.a.values)
You get
array([[ 0, 0, 0],
[ 1, 2, 3],
[ 2, 4, 6],
[ 3, 6, 9],
[ 4, 8, 12],
[ 5, 10, 15],
[ 6, 12, 18],
[ 7, 14, 21],
[ 8, 16, 24],
[ 9, 18, 27]])
Check
np.array(df['a'].tolist())
array([[ 0, 0, 0],
[ 1, 2, 3],
[ 2, 4, 6],
[ 3, 6, 9],
[ 4, 8, 12],
[ 5, 10, 15],
[ 6, 12, 18],
[ 7, 14, 21],
[ 8, 16, 24],
[ 9, 18, 27]], dtype=int64)
I have a data set like this
a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
How can I reshape this into shape (3,2,2) so that a[:,0,0] = [1,2,3]?
you can use two steps:
step1.
In [28]: b1 = np.reshape(a,(3,4), order='F')
In [29]: b1
Out[29]:
array([[ 1, 4, 7, 10],
[ 2, 5, 8, 11],
[ 3, 6, 9, 12]])
use order='F' means to read / write the elements using Fortran-like index order, with the first index changing fastest, and the last index changing slowest. numpy.reshape
setp2
In [30]: c = b1.reshape(3,2,2)
In [31]: c
Out[31]:
array([[[ 1, 4],
[ 7, 10]],
[[ 2, 5],
[ 8, 11]],
[[ 3, 6],
[ 9, 12]]])
get the final result:
In [34]: c[:,0,0]
Out[34]: array([1, 2, 3])
In [30]: a=np.arange(1,13)
In [31]: a
Out[31]: array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12])
Since you want to keep the first 3 values 'together', we could start with a reshape like:
In [32]: a.reshape(2,2,3)
Out[32]:
array([[[ 1, 2, 3],
[ 4, 5, 6]],
[[ 7, 8, 9],
[10, 11, 12]]])
and then swap a couple of the axes:
In [33]: a.reshape(2,2,3).transpose(2,0,1)
Out[33]:
array([[[ 1, 4],
[ 7, 10]],
[[ 2, 5],
[ 8, 11]],
[[ 3, 6],
[ 9, 12]]])
In [34]: _[:,0,0]
Out[34]: array([1, 2, 3])
Or with a different transpose:
In [35]: a.reshape(2,2,3).transpose(2,1,0)
Out[35]:
array([[[ 1, 7],
[ 4, 10]],
[[ 2, 8],
[ 5, 11]],
[[ 3, 9],
[ 6, 12]]])
transpose() with an argument, (also invoked with .T) does the same thing.
So your question is a bit ambiguous.
So does the reshape with order F mentioned in the other answer:
In [37]: a.reshape(3,2,2, order='F')
Out[37]:
array([[[ 1, 7],
[ 4, 10]],
[[ 2, 8],
[ 5, 11]],
[[ 3, 9],
[ 6, 12]]])
(though the two step, a.reshape(3,4, order='F').reshape(3,2,2) produces my first result Out[33]).
Given this array:
>>> a
array([[ 0, 1, 2],
[ 3, 4, 5],
[ 6, 7, 8],
[ 9, 10, 11]])
How can I select [[4,5], [7,8]]? a[0::2, 1:;2] doesn't work
>>> a
array([[ 0, 1, 2],
[ 3, 4, 5],
[ 6, 7, 8],
[ 9, 10, 11]])
>>> a[1:3,1:3]
array([[4, 5],
[7, 8]])
The first 1:3 is to select row 1 & 2. The second 1:3 is to select column 1 & 2.
I have an array :
e = np.array([[ 0, 1, 2, 3, 5, 6, 7, 8],
[ 4, 5, 6, 7, 5, 3, 2, 5],
[ 8, 9, 10, 11, 4, 5, 3, 5]])
I want to extract array by its columns in RANGE, if I want to take column in range 1 until 5, It will return
e = np.array([[ 1, 2, 3, 5, ],
[ 5, 6, 7, 5, ],
[ 9, 10, 11, 4, ]])
How to solve it? Thanks
You can just use e[:, 1:5] to retrive what you want.
In [1]: import numpy as np
In [2]: e = np.array([[ 0, 1, 2, 3, 5, 6, 7, 8],
...: [ 4, 5, 6, 7, 5, 3, 2, 5],
...: [ 8, 9, 10, 11, 4, 5, 3, 5]])
In [3]: e[:, 1:5]
Out[3]:
array([[ 1, 2, 3, 5],
[ 5, 6, 7, 5],
[ 9, 10, 11, 4]])
https://docs.scipy.org/doc/numpy/reference/arrays.indexing.html
Numpy row and column indices start counting at 0.
The rows are specified first and then the column with a comma to separate the row from column.
The ":" (colon) is used to shortcut all rows or all columns when it is used alone.
When the row or column specifier has a range, then the ":" is paired with numbers that specify the inclusive start range and the exclusive end range.
For example
import numpy as np
np_array = np.array( [ [ 1, 2, 3, ],
[ 4, 5, 6, ],
[ 7, 8, 9 ] ] )
first_row = np_array[0,:]
first_row
output: array([1, 2, 3])
last_column = np_array[:,2]
last_column
output: array([3, 6, 9])
first_two_vals = np_array[0,0:2]
first_two_vals
output: array([1, 2])
You can use np.take with specifying axis=1
import numpy as np
e = np.array([[ 0, 1, 2, 3, 5, 6, 7, 8],
[ 4, 5, 6, 7, 5, 3, 2, 5],
[ 8, 9, 10, 11, 4, 5, 3, 5]])
e = np.take(e, [1,2,3,4], axis=1)
output:
array([[ 1, 2, 3, 5],
[ 5, 6, 7, 5],
[ 9, 10, 11, 4]])
https://numpy.org/doc/stable/reference/generated/numpy.take.html
I am wondering if there is a way it index/slice a numpy array, such that one can get every other band of 2 elements. In other words, given:
test = np.array([[1,2,3,4,5,6,7,8],[9,10,11,12,13,14,15,16]])
I would like to get the array:
[[1, 2, 5, 6],
[9, 10, 13, 14]]
Thoughts on how this can be accomplished with slicing/indexing?
Not that difficult with a few smart reshapes :)
test.reshape((4, 4))[:, :2].reshape((2, 4))
Given:
>>> test
array([[ 1, 2, 3, 4, 5, 6, 7, 8],
[ 9, 10, 11, 12, 13, 14, 15, 16]])
You can do:
>>> test.reshape(-1,2)[::2].reshape(-1,4)
array([[ 1, 2, 5, 6],
[ 9, 10, 13, 14]])
Which works even for different shapes of initial arrays:
>>> test2
array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16])
>>> test2.reshape(-1,2)[::2].reshape(-1,4)
array([[ 1, 2, 5, 6],
[ 9, 10, 13, 14]])
>>> test3
array([[ 1, 2, 3, 4],
[ 5, 6, 7, 8],
[ 9, 10, 11, 12],
[13, 14, 15, 16]])
>>> test3.reshape(-1,2)[::2].reshape(-1,4)
array([[ 1, 2, 5, 6],
[ 9, 10, 13, 14]])
How it works:
1. Reshape into two columns by however many rows:
>>> test.reshape(-1,2)
array([[ 1, 2],
[ 3, 4],
[ 5, 6],
[ 7, 8],
[ 9, 10],
[11, 12],
[13, 14],
[15, 16]])
2. Stride the array by stepping every second element
>>> test.reshape(-1,2)[::2]
array([[ 1, 2],
[ 5, 6],
[ 9, 10],
[13, 14]])
3. Set the shape you want of 4 columns, however many rows:
>>> test.reshape(-1,2)[::2].reshape(-1,4)
array([[ 1, 2, 5, 6],
[ 9, 10, 13, 14]])