将列表转换并填充为 numpy 数组 [英] Convert and pad a list to numpy array
问题描述
我有一个任意深度嵌套的列表,具有不同长度的元素
my_list = [[[1,2],[4]],[[4,4,3]],[[1,2,1],[4,3,4,5],[4,1]]]
我想通过用 NaN 填充每个轴,将其转换为有效的数字(非对象)numpy 数组.所以结果应该是这样的
padded_list = np.array([[[ 1, 2, nan, nan],[ 4, 难, 难, 难],[南,南,南,南]],[[ 4, 4, 3, 南],[楠,楠,楠,楠],[南,南,南,南]],[[ 1, 2, 1, 南],[ 4, 3, 4, 5],[ 4, 1, 南, 南]]])
我该怎么做?
这适用于您的示例,不确定它是否可以正确处理所有极端情况:
from itertools import izip_longestdef find_shape(seq):尝试:len_ = len(seq)除了类型错误:返回 ()形状 = [find_shape(subseq) for subseq in seq]返回 (len_,) + tuple(max(sizes) for size in izip_longest(*shapes,填充值=1))def fill_array(arr, seq):如果 arr.ndim == 1:尝试:len_ = len(seq)除了类型错误:len_ = 0arr[:len_] = seqarr[len_:] = np.nan别的:对于 izip_longest(arr, seq, fillvalue=()) 中的 subarr, subseq:填充数组(subarr,subseq)
现在:
<预><代码>>>>arr = np.empty(find_shape(my_list))>>>填充数组(arr,my_list)>>>阿尔数组([[[ 1., 2., nan, nan],[ 4., 楠, 楠, 楠],[楠,楠,楠,楠]],[[4., 4., 3., 南],[楠,楠,楠,楠],[楠,楠,楠,楠]],[[ 1., 2., 1., 南],[ 4., 3., 4., 5.],[ 4., 1., nan, nan]]]])我认为这大致就是 numpy 的形状发现例程所做的.由于无论如何都涉及大量 Python 函数调用,因此与 C 实现相比,它可能不会那么糟糕.
I have an arbitrarily deeply nested list, with varying length of elements
my_list = [[[1,2],[4]],[[4,4,3]],[[1,2,1],[4,3,4,5],[4,1]]]
I want to convert this to a valid numeric (not object) numpy array, by padding out each axis with NaN. So the result should look like
padded_list = np.array([[[ 1, 2, nan, nan],
[ 4, nan, nan, nan],
[nan, nan, nan, nan]],
[[ 4, 4, 3, nan],
[nan, nan, nan, nan],
[nan, nan, nan, nan]],
[[ 1, 2, 1, nan],
[ 4, 3, 4, 5],
[ 4, 1, nan, nan]]])
How do I do this?
This works on your sample, not sure it can handle all the corner cases properly:
from itertools import izip_longest
def find_shape(seq):
try:
len_ = len(seq)
except TypeError:
return ()
shapes = [find_shape(subseq) for subseq in seq]
return (len_,) + tuple(max(sizes) for sizes in izip_longest(*shapes,
fillvalue=1))
def fill_array(arr, seq):
if arr.ndim == 1:
try:
len_ = len(seq)
except TypeError:
len_ = 0
arr[:len_] = seq
arr[len_:] = np.nan
else:
for subarr, subseq in izip_longest(arr, seq, fillvalue=()):
fill_array(subarr, subseq)
And now:
>>> arr = np.empty(find_shape(my_list))
>>> fill_array(arr, my_list)
>>> arr
array([[[ 1., 2., nan, nan],
[ 4., nan, nan, nan],
[ nan, nan, nan, nan]],
[[ 4., 4., 3., nan],
[ nan, nan, nan, nan],
[ nan, nan, nan, nan]],
[[ 1., 2., 1., nan],
[ 4., 3., 4., 5.],
[ 4., 1., nan, nan]]])
I think this is roughly what the shape discovery routines of numpy do. Since there are lots of Python function calls involved anyway, it probably won't compare that badly against the C implementation.
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