如何将带有空 arrays 的 numpy arrays 列表转换为单个数组？

Question

So my data looks like:

[array([0], dtype=int64),
 array([1], dtype=int64),
 array([1], dtype=int64),
 array([2], dtype=int64),
 array([3], dtype=int64),
 array([3], dtype=int64),
 array([4], dtype=int64),
 array([], dtype=int64),
 array([], dtype=int64),
 array([], dtype=int64),
 array([], dtype=int64),
 array([], dtype=int64),
 array([], dtype=int64),
 array([], dtype=int64),
 array([], dtype=int64),
 array([], dtype=int64),
 array([], dtype=int64),
 array([], dtype=int64),
 array([], dtype=int64),
 array([], dtype=int64),
 array([6], dtype=int64) ...

While using np.concatenate(list_1) does concatenate the arrays but skips the empty arrays. And in tthe resultant array 6 is the next element to 4 and intermediate empty arrays does not appear in the list. This is what np.concateneate does but I do not want it that way.

I want to combine those arrays into a single array having same length as the list but NaN values in place of empty arrays. How can I achieve that?

Answer 1

One way:

np.concatenate([a if a.size else np.array([np.nan]) for a in array_list])

I would hazard a guess though there is probably a better way to load your data.

As a side note, concatenate does not skip any array - it concatenates the arrays, that is, puts the elements for each one after the other. Placing 0 elements is just not noticeable.

Answer 2

You can pre-allocate the output and assign to it directly:

result = np.full(len(list_1), np.nan)
for i, v in enumerate(list_1):
    if v.size:
        result[i] = v.item()

Alternatively, you can do

result = np.empty(len(list_1))
for i, v in enumerate(list_1):
    result[i] = v.item() if v.size else np.nan