Sqrt only one column in numpy-CodePudding

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]])

CodePudding user response：

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

CodePudding user response：

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.

CodePudding user response：

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)