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I'm doing a TensorFlow tutorial, where they convert an array of the numbers [1,2,3] to a tensor like this:

const xs = tf.tensor2d([1, 2, 3], [3, 1])

The shape is [3,1] because there is one row with 3 numbers. My question is, why would they use a 2D tensor, isn't this just exactly the same as:

const xs = tf.tensor1d([1, 2, 3])
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  • $\begingroup$ This seems to be just a programming question/issue. Am I right? If yes, then you should ask this question on Stack Overflow, as programming questions are generally off-topic here. Please, if you have some time, take a look at our on-topic page, where we describe what kind of question you can ask here. $\endgroup$
    – nbro
    Commented Apr 16, 2021 at 9:53
  • $\begingroup$ I posted it here because it's not really about code but about the difference in 1d or 2d tensors. Unless that is also considered code? $\endgroup$
    – Kokodoko
    Commented Apr 16, 2021 at 10:17
  • $\begingroup$ To give a proper answer, I think we will need more context, i.e. please provide the link to the tutorial where the first version of the code is used. $\endgroup$
    – nbro
    Commented Apr 16, 2021 at 10:53

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The required shape of the tensor $T$ depends on the shape of other tensors that are involved in the same operations of that same tensor $T$ and the required/desired shape of the resulting tensor, in the same way that the number of columns of the matrix $M \in \mathbb{R}^{n \times m}$ needs to match the number of rows of the matrix $M' \in \mathbb{R}^{n' \times m'}$ when you perform the matrix multiplication $M M'$, i.e. in order for $M M'$ to be well-defined, $m = n'$.

So, in your case, although the tensors contain the same elements, it might not be possible to use both in the same operations. Without more context/details, I cannot specifically answer why the first tensor was used in the tutorial you're mentioning.

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