Numpy index slice without losing dimension information

user_339573 0
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I'm using numpy and want to index a row without losing the dimension information.

import numpy as np
X = np.zeros((100,10))
X.shape        # >> (100, 10)
xslice = X[10,:]
xslice.shape   # >> (10,)

In this example xslice is now 1 dimension, but I want it to be (1,10). In R, I would use X[10,:,drop=F]. Is there something similar in numpy. I couldn't find it in the documentation and didn't see a similar question asked.

Thanks!

numpy python
7 Answers

Here's an alternative I like better. Instead of indexing with a single number, index with a range. That is, use X[10:11,:]. (Note that 10:11 does not include 11).

import numpy as np
X = np.zeros((100,10))
X.shape        # >> (100, 10)
xslice = X[10:11,:]
xslice.shape   # >> (1,10)

This makes it easy to understand with more dimensions too, no None juggling and figuring out which axis to use which index. Also no need to do extra bookkeeping regarding array size, just i:i+1 for any i that you would have used in regular indexing.

b = np.ones((2, 3, 4))
b.shape # >> (2, 3, 4)
b[1:2,:,:].shape  # >> (1, 3, 4)
b[:, 2:3, :].shape .  # >> (2, 1, 4)
user_21025910
user_18391060

This is especially annoying if you're indexing by an array that might be length 1 at runtime. For that case, there's np.ix_: 

some_array[np.ix_(row_index,column_index)]
user_15191990

I've been using np.reshape to achieve the same as shown below

import numpy as np
X = np.zeros((100,10))
X.shape        # >> (100, 10)
xslice = X[10,:].reshape(1, -1)
xslice.shape   # >> (1, 10)
user_60026240
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