将具有不同大小行的数据加载到 Numpy 数组中 [英] Load data with rows of different sizes into Numpy array

问题描述

假设我有一个包含如下数据的文本文件:

Suppose I have a text file that contains data like this:

1  2  3  4  5
6  7  8 
9  10 11 12 13 14
15 16 17 18 19

如何将其加载到 numpy 数组中，使其看起来像这样?

How do I load it into a numpy array so it looks like this?

[1  2  3  4  5  0
 6  7  8  0  0  0
 9  10 11 12 13 14
 15 16 17 18 19 0 ]

到目前为止我一直在使用的方法是逐行读取文本文件，将每一行附加到一个列表中，找到长度最大的行并相应地填充剩余的行.

The method I've been using so far involves reading the text file line by line, appending each row to a list, finding the row with the maximum length and padding the remaining rows accordingly.

谁能提出更有效的方法?

Could anyone suggest a more efficient way?

非常感谢！

推荐答案

可以通过多种方式填充列表列表，但是由于您已经从文件中读取了此内容，我认为 itertools.zip_longest 将是一个好的开始.

Padding a list of lists can be done in various ways, but since you are already reading this from a file, I think the itertools.zip_longest will be a good start.

In [201]: txt = """1  2  3  4  5
     ...: 6  7  8 
     ...: 9  10 11 12 13 14
     ...: 15 16 17 18 19"""

读取并解析文本行:

In [202]: alist = []
In [203]: for line in txt.splitlines():
     ...:     alist.append([int(i) for i in line.split()])
     ...:     
In [204]: alist
Out[204]: [[1, 2, 3, 4, 5], [6, 7, 8], [9, 10, 11, 12, 13, 14], [15, 16, 17, 18, 19]]

zip_longest(此处为 PY3 形式)采用填充值:

zip_longest (here in PY3 form) takes a fillvalue:

In [205]: from itertools import zip_longest
In [206]: list(zip_longest(*alist, fillvalue=0))
Out[206]: 
[(1, 6, 9, 15),
 (2, 7, 10, 16),
 (3, 8, 11, 17),
 (4, 0, 12, 18),
 (5, 0, 13, 19),
 (0, 0, 14, 0)]
In [207]: np.array(_).T
Out[207]: 
array([[ 1,  2,  3,  4,  5,  0],
       [ 6,  7,  8,  0,  0,  0],
       [ 9, 10, 11, 12, 13, 14],
       [15, 16, 17, 18, 19,  0]])

zip(*) 也可用于转置"列表列表:

zip(*) can also be used to 'transpose' the list of lists:

In [209]: list(zip(*alist1))
Out[209]: 
[(1, 2, 3, 4, 5, 0),
 (6, 7, 8, 0, 0, 0),
 (9, 10, 11, 12, 13, 14),
 (15, 16, 17, 18, 19, 0)]

我猜你正在做的事情更像是:

I'm guessing you are doing something more like:

In [211]: maxlen = max([len(i) for i in alist])
In [212]: maxlen
Out[212]: 6
In [213]: arr = np.zeros((len(alist), maxlen),int)
In [214]: for row, line in zip(arr, alist):
     ...:     row[:len(line)] = line
     ...:     
In [215]: arr
Out[215]: 
array([[ 1,  2,  3,  4,  5,  0],
       [ 6,  7,  8,  0,  0,  0],
       [ 9, 10, 11, 12, 13, 14],
       [15, 16, 17, 18, 19,  0]])

这对我来说看起来不错.

Which looks pretty good to me.

一个普通的发帖人 Divakar 喜欢发布一个使用 cumsum 的解决方案.看看我能不能重现它.它涉及构建一个非零值应该去的一维掩码.向后工作我们需要一个面具，如:

A regular poster, Divakar, likes to post a solution that uses cumsum. Let's see if I can reproduce it. It involves constructing a 1d mask where the nonzero values are supposed to go. Working backwards we need a mask like:

In [240]: mask=arr.ravel()>0
In [241]: mask
Out[241]: 
array([ True,  True,  True,  True,  True, False,  True,  True,  True,
       False, False, False,  True,  True,  True,  True,  True,  True,
        True,  True,  True,  True,  True, False], dtype=bool)

所以:

In [242]: arr.flat[mask] = np.hstack(alist)

这个映射有一个我还没有完全内化的技巧！

There's a trick to this mapping that I haven't quite internalized!

诀窍是针对 [0,1,2,3,4,5] 广播长度:

The trick is the broadcast the lengths against [0,1,2,3,4,5]:

In [276]: lens=[len(i) for i in alist]
In [277]: maxlen=max(lens)
In [278]: mask=np.array(lens)[:,None]>np.arange(maxlen)
In [279]: mask
Out[279]: 
array([[ True,  True,  True,  True,  True, False],
       [ True,  True,  True, False, False, False],
       [ True,  True,  True,  True,  True,  True],
       [ True,  True,  True,  True,  True, False]], dtype=bool)
In [280]: arr = np.zeros((len(alist), maxlen),int)
In [281]: arr[mask] = np.hstack(alist)
In [282]: arr
Out[282]: 
array([[ 1,  2,  3,  4,  5,  0],
       [ 6,  7,  8,  0,  0,  0],
       [ 9, 10, 11, 12, 13, 14],
       [15, 16, 17, 18, 19,  0]])

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