How to replace only 1d values in 2d array after filter using numpy in python without loop i.e in pythonic way.

I want to filter only t2 rows and replace values in second column ( middle column ).

example:

`>>> x = np.array([['t1',10,20],['t2',11,22],['t2',12,23], ['t3',21,32]])`

``````>>> x
array([['t1', '10', '20'],
['t2', '11', '22'],
['t2', '12', '23'],
['t3', '21', '32']],
dtype='|S2')
``````

replace_value_for_t2_col1 = [100,101]

so expected output is
``````array([['t1', '10', '20'],
['t2', '100', '22'],
['t2', '101', '23'],
['t3', '21', '32']],
dtype='|S2')
``````
What i tried was,

Filter rows for t2

`>>> x[x[:,0]=='t2']`

``````array([['t2', '11', '22'],
['t2', '12', '23']],
dtype='|S2')
``````

`>>> x[x[:,0]=='t2'][:,1] = np.array([101,102])`

Seems right, but not replacing. i guess it make a copy when we slice using [:,1], so its not changing in same array.

Please suggest me any brilliant ,simple and pythonic ideas, techniques.

`x[x[:,0]=='t2']`
gives you a new array

The pythonic way would be to use a loop.

I would just temporarily convert back to a Python list:

``````import numpy as np

x = np.array([['t1',10,20],['t2',11,22],['t2',12,23], ['t3',21,32]])
print(x)  # test

# convert temporarily to Python list again
mylist = x.tolist()
mylist2 = ['100', '101']
n = 0
newlist = []
for line in mylist:
if line[0] == 't2':
line[1] = mylist2[n]
n += 1
newlist.append(line)

print(newlist)  # test

# convert back to numpy array
x = np.array(newlist)
print(x)  # show result
``````
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