I have the following 2D array, and want to take a square root of only column A.
import numpy as np
a = np.matrix([[1, 2], [3, 4], [5, 6], [7, 8]])
a
matrix([[1, 2],
[3, 4],
[5, 6],
[7, 8]])
This is giving me sqrt of two columns. How can I only take a square root of column A?
b = np.sqrt(a[:, [0, 1]])
b
matrix([[1. , 1.41421356],
[1.73205081, 2. ],
[2.23606798, 2.44948974],
[2.64575131, 2.82842712]])
Use out to do in place operation
import numpy as np
a = np.matrix([[1, 2], [3, 4], [5, 6], [7, 8]], dtype=np.float64)
np.sqrt(a, where=[True, False],out=a)
Output:
[[1. 2. ]
[1.73205081 4. ]
[2.23606798 6. ]
[2.64575131 8. ]]
Try it online
I can't comment so I'm just going to have to put my answer here:
You can do it in-place with the following:
a = np.matrix([[1, 2], [3, 4], [5, 6], [7, 8]], dtype=np.float)
a[:, 0] = np.sqrt(a[:, 0])
Do note that because your initial types were int, that you must specify the dtype=np.float if not it will all get cast to ints.
if you want only first column, you must this
b = np.sqrt(a[:, [0]])
if you want all columns but first columns sqrt, you try this
df=pd.DataFrame(a)
df.loc[:,0]=df.loc[:,0].apply(np.sqrt)
Let say I have 2 numpy arrays
import numpy as np
x = np.array([1,2,3])
y = np.array([1,2,3,4])
With this, I want to create a 2-dimensional array as below
Is there any method available to directly achieve this?
You problem is about writing the Cartesian product. In numpy, you can write it using repeat and tile:
out = np.c_[np.repeat(x, len(y)), np.tile(y, len(x))]
Python's builtin itertools module has a method designed for this: product:
from itertools import product
out = np.array(list(product(x,y)))
Output:
array([[1, 1],
[1, 2],
[1, 3],
[1, 4],
[2, 1],
[2, 2],
[2, 3],
[2, 4],
[3, 1],
[3, 2],
[3, 3],
[3, 4]])
I am learning Python and solving a machine learning problem.
class_ids=np.arange(self.x.shape[0])
np.random.shuffle(class_ids)
self.x=self.x[class_ids]
This is a shuffle function in NumPy but I can't understand what self.x=self.x[class_ids] means. because I think it gives the value of the array to a variable.
It's a very complicated way to shuffle the first dimension of your self.x. For example:
>>> x = np.array([[1, 1], [2, 2], [3, 3], [4, 4], [5, 5]])
>>> x
array([[1, 1],
[2, 2],
[3, 3],
[4, 4],
[5, 5]])
Then using the mentioned approach
>>> class_ids=np.arange(x.shape[0]) # create an array [0, 1, 2, 3, 4]
>>> np.random.shuffle(class_ids) # shuffle the array
>>> x[class_ids] # use integer array indexing to shuffle x
array([[5, 5],
[3, 3],
[1, 1],
[4, 4],
[2, 2]])
Note that the same could be achieved just by using np.random.shuffle because the docstring explicitly mentions:
This function only shuffles the array along the first axis of a multi-dimensional array. The order of sub-arrays is changed but their contents remains the same.
>>> np.random.shuffle(x)
>>> x
array([[5, 5],
[3, 3],
[1, 1],
[2, 2],
[4, 4]])
or by using np.random.permutation:
>>> class_ids = np.random.permutation(x.shape[0]) # shuffle the first dimensions indices
>>> x[class_ids]
array([[2, 2],
[4, 4],
[3, 3],
[5, 5],
[1, 1]])
Assuming self.x is a numpy array:
class_ids is a 1-d numpy array that is being used as an integer array index in the expression: x[class_ids]. Because the previous line shuffled class_ids, x[class_ids] evaluates to self.x shuffled by rows.
The assignment self.x=self.x[class_ids] assigns the shuffled array to self.x
Seemingly simple question: I have an array with two columns, the first represents an ID and the second a count. I'd like to update it with another, similar array such that
import numpy as np
a = np.array([[1, 2],
[2, 2],
[3, 1],
[4, 5]])
b = np.array([[2, 2],
[3, 1],
[4, 0],
[5, 3]])
a.update(b) # ????
>>> np.array([[1, 2],
[2, 4],
[3, 2],
[4, 5],
[5, 3]])
Is there a way to do this with indexing/slicing such that I don't simply have to iterate over each row?
Generic case
Approach #1: You can use np.add.at to do such an ID-based adding operation like so -
# First column of output array as the union of first columns of a,b
out_id = np.union1d(a[:,0],b[:,0])
# Initialize second column of output array
out_count = np.zeros_like(out_id)
# Find indices where the first columns of a,b are placed in out_id
_,a_idx = np.where(a[:,None,0]==out_id)
_,b_idx = np.where(b[:,None,0]==out_id)
# Place second column of a into out_id & add in second column of b
out_count[a_idx] = a[:,1]
np.add.at(out_count, b_idx,b[:,1])
# Stack the ID and count arrays into a 2-column format
out = np.column_stack((out_id,out_count))
To find a_idx and b_idx, as probably a faster alternative, np.searchsorted could be used like so -
a_idx = np.searchsorted(out_id, a[:,0], side='left')
b_idx = np.searchsorted(out_id, b[:,0], side='left')
Sample input-output :
In [538]: a
Out[538]:
array([[1, 2],
[4, 2],
[3, 1],
[5, 5]])
In [539]: b
Out[539]:
array([[3, 7],
[1, 1],
[4, 0],
[2, 3],
[6, 2]])
In [540]: out
Out[540]:
array([[1, 3],
[2, 3],
[3, 8],
[4, 2],
[5, 5],
[6, 2]])
Approach #2: You can use np.bincount to do the same ID based adding -
# First column of output array as the union of first columns of a,b
out_id = np.union1d(a[:,0],b[:,0])
# Get all IDs and counts in a single arrays
id_arr = np.concatenate((a[:,0],b[:,0]))
count_arr = np.concatenate((a[:,1],b[:,1]))
# Get binned summations
summed_vals = np.bincount(id_arr,count_arr)
# Get mask of valid bins
mask = np.in1d(np.arange(np.max(out_id)+1),out_id)
# Mask valid summed bins for final counts array output
out_count = summed_vals[mask]
# Stack the ID and count arrays into a 2-column format
out = np.column_stack((out_id,out_count))
Specific case
If the ID columns in a and b are sorted, it becomes easier, as we can just use masks with np.in1d to index into the output ID array created with np.union like so -
# First column of output array as the union of first columns of a,b
out_id = np.union1d(a[:,0],b[:,0])
# Masks of first columns of a and b matches in the output ID array
mask1 = np.in1d(out_id,a[:,0])
mask2 = np.in1d(out_id,b[:,0])
# Initialize second column of output array
out_count = np.zeros_like(out_id)
# Place second column of a into out_id & add in second column of b
out_count[mask1] = a[:,1]
np.add.at(out_count, np.where(mask2)[0],b[:,1])
# Stack the ID and count arrays into a 2-column format
out = np.column_stack((out_id,out_count))
Sample run -
In [552]: a
Out[552]:
array([[1, 2],
[2, 2],
[3, 1],
[4, 5],
[8, 5]])
In [553]: b
Out[553]:
array([[2, 2],
[3, 1],
[4, 0],
[5, 3],
[6, 2],
[8, 2]])
In [554]: out
Out[554]:
array([[1, 2],
[2, 4],
[3, 2],
[4, 5],
[5, 3],
[6, 2],
[8, 7]])
>>> col=np.unique(np.hstack((b[:,0],a[:,0])))
>>> dif=np.setdiff1d(col,a[:,0])
>>> val=b[np.in1d(b[:,0],dif)]
>>> result=np.concatenate((a,val))
array([[1, 2],
[2, 2],
[3, 1],
[4, 5],
[5, 3]])
Note that if you want the result become sorted you can use np.lexsort :
result[np.lexsort((result[:,0],result[:,0]))]
Explanation :
First you can find the unique ids with following command :
>>> col=np.unique(np.hstack((b[:,0],a[:,0])))
>>> col
array([1, 2, 3, 4, 5])
Then find the different between the ids if a and all of ids :
>>> dif=np.setdiff1d(col,a[:,0])
>>> dif
array([5])
Then find the items within b with the ids in diff :
>>> val=b[np.in1d(b[:,0],dif)]
>>> val
array([[5, 3]])
And at last concatenate the result with list a:
>>> np.concatenate((a,val))
consider another example with sorting :
>>> a = np.array([[1, 2],
... [2, 2],
... [3, 1],
... [7, 5]])
>>>
>>> b = np.array([[2, 2],
... [3, 1],
... [4, 0],
... [5, 3]])
>>>
>>> col=np.unique(np.hstack((b[:,0],a[:,0])))
>>> dif=np.setdiff1d(col,a[:,0])
>>> val=b[np.in1d(b[:,0],dif)]
>>> result=np.concatenate((a,val))
>>> result[np.lexsort((result[:,0],result[:,0]))]
array([[1, 2],
[2, 2],
[3, 1],
[4, 0],
[5, 3],
[7, 5]])
That's an old question but here is a solution with pandas (that could be generalized for other aggregation functions than sum). Also sorting will occur automatically:
import pandas as pd
import numpy as np
a = np.array([[1, 2],
[2, 2],
[3, 1],
[4, 5]])
b = np.array([[2, 2],
[3, 1],
[4, 0],
[5, 3]])
print((pd.DataFrame(a[:, 1], index=a[:, 0])
.add(pd.DataFrame(b[:, 1], index=b[:, 0]), fill_value=0)
.astype(int))
.reset_index()
.to_numpy())
Output:
[[1 2]
[2 4]
[3 2]
[4 5]
[5 3]]