What is Numpy equivalence of dataframe.loc() in Pandas

Question

I have a 120,000*4 numpy array as shown below. Each row is a sample. The first column is time in second, or the index using Pandas terminology.

0.014      14.175  -29.97  -22.68 
0.022      13.905  -29.835 -22.68
0.030      12.257  -29.32  -22.67
... ...
1259.980   -0.405   2.205   3.825
1259.991   -0.495   2.115   3.735

I want to select the rows recorded between 100.000 to 200.000 sec and save it into a new array. If this were a Pandas dataframe, I would simply write df.loc[100:200]. What is the equivalent operation in numpy?

This is NOT a question of feasibility. I simply wonder if there are any pythonic one-line solutions.

Answer 1

This assumes indexes are sorted:

IIUC,

x=np.array([ [1,2,3,4],
           [5,6,7,8],
           [9,10,11,12],
           [13,14,15,16]])

x[(x[:,0] >= 5) & (x[:,0] <= 9) ]

So you would have 100 and 200 instead of 5 and 9.

---

For a more general solution, check Wen`s answer

Answer 2

Data from Raf

x[np.where(x[:,0]==5)[0][0]:np.where(x[:,0]==9)[0][0]+1,:]
Out[341]: 
array([[ 5,  6,  7,  8],
       [ 9, 10, 11, 12]])

Notice

only using greater and less than for that can not fully replace the .loc, the back end of .loc is index position not value range

For example

df
Out[348]: 
       0   1   2   3
0      1   2   3   4
1      5   6   7   8
4444   9  10  11  12
3     13  14  15  16

df.loc[1:3]
Out[347]: 
       0   1   2   3
1      5   6   7   8
4444   9  10  11  12
3     13  14  15  16