I have a numpy array such as arr = np.arange(16).reshape(2,2,2,2)
I want to dynamically access arr[:, dim2, dim3], when I have (dim2, dim3) as a tuple. What is the best way to do this?
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Are you thinking of a case where you know the dimensions of your array in advance, or are you trying to think of a method that would work even if you didn't know, say, the number of dimensions in the array?jjramsey– jjramsey2020-02-27 13:35:53 +00:00Commented Feb 27, 2020 at 13:35
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@jjramsey The latter, I'll make that clearer.Jack Black– Jack Black2020-02-27 13:37:58 +00:00Commented Feb 27, 2020 at 13:37
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2 Answers
Try something like this if the number of dimensions might not be the same for your array:
some_tuple = (dim2, dim3) # Could be (dim2, dim3, ..., dimN)
arr[(slice(None),) + some_tuple]
In this particular case, (slice(None),) + some_tuple is the same as (slice(None), dim2, dim3). slice(None) is more or less equivalent to ":", but it can be used in more places than ":". Notice that I put slice(None) in a single-element tuple (i.e. (slice(None),)) so that I can add it to some_tuple. Notice as well that there's a comma after slice(None), i.e., I don't just write (slice(None)) without a comma. It won't work without the extra comma.
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yatu
Nice answer. Alternatively
arr[(slice(None), *some_tuple)] could also be an option, this way you avoid creating a tuplesome_tuple = (dim2, dim3)
arr[:, some_tuple[0], some_tuple[1]]
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Jack Black
Thanks! Is there a general way to do this, especially if I have a long tuple?