英文:
Why full function in NumPy can't take dtype=str
问题
我为您提供翻译后的部分，不包括代码：
为什么我看到的只有一个字母而不是“cube”？
当我没有输入dtype="str"时，它正常工作。
我的问题是为什么？
结果：
array([[['c', 'c', 'c'],
['c', 'c', 'c'],
['c', 'c', 'c']],
[['c', 'c', 'c'],
['c', 'c', 'c'],
['c', 'c', 'c']],
[['c', 'c', 'c'],
['c', 'c', 'c'],
['c', 'c', 'c']]], dtype='<U1')
英文:
Why i see just one letter instead of "cube"
When i didn't type dtype="str" it worked.
My question is why?
code:
np.full((3,3,3),"cube",dtype=str)
results:
array([[['c', 'c', 'c'],
['c', 'c', 'c'],
['c', 'c', 'c']],
[['c', 'c', 'c'],
['c', 'c', 'c'],
['c', 'c', 'c']],
[['c', 'c', 'c'],
['c', 'c', 'c'],
['c', 'c', 'c']]], dtype='<U1')
答案1
得分: 1
np.full((3,3,3),"cube",dtype=str)
等同于
np.full((3,3,3),"cube",dtype='U1')
它将只取字符串的第一个元素。
您可以通过以下方式获得完整的结果：
np.full((3,3,3),"cube",dtype=object)
或者
np.full((3,3,3),"cube",dtype='U4') # 'U4'表示cube有4个字母，您也可以使用U>4，比如U5、U6等。
或者只需移除 dtype=str。移除后将取整个字符串的长度。
np.full((3,3,3),"cube")
英文:
np.full((3,3,3),"cube",dtype=str)
is equivalent to
np.full((3,3,3),"cube",dtype='U1')
It will just take the first element of the string.
You can get full result by:
np.full((3,3,3),"cube",dtype=object)
or
np.full((3,3,3),"cube",dtype='U4') # U4 as you have 4 letters in cube, you can do any U>4 like U5, U6, etc.
or just remove dtype=str. Upon removing it will take the complete length of the string.
np.full((3,3,3),"cube")
答案2
得分: 1
以下是要翻译的代码部分：
if dtype is None:
fill_value = asarray(fill_value)
dtype = fill_value.dtype
a = empty(shape, dtype, order)
multiarray.copyto(a, fill_value, casting='unsafe')
np.asarray 可以推断出 cube 的填充值的 U4 数据类型。
但是 np.empty(shape, 'str') 生成了 U1 数据类型。它对于 fill_value 没有关于数据类型的信息。
所以，这就是为什么在没有指定数据类型的情况下，它会推断需要足够大的字符串来容纳4个字符，但是没有指定数据类型时，它只分配了1个字符的空间。
英文:
Part of the source code for full is (see the [source] link in docs):
if dtype is None:
fill_value = asarray(fill_value)
dtype = fill_value.dtype
a = empty(shape, dtype, order)
multiarray.copyto(a, fill_value, casting='unsafe')
np.asarray can deduce a 'U4' dtype for the fill value of 'cube'.
But np.empty(shape, 'str') produces a 'U1' dtype. It has no information about fill_value.
So that's why, without your dtype, it deduces that it needs a large enough string to hold 4 characters, but without it, it just allocates space for 1 character.
答案3
得分: 0
那应该可以运行
import numpy as np
np.full((3,3,3),"cube",dtype='U4')
英文:
That should work
import numpy as np
np.full((3,3,3),"cube",dtype='U4')
答案4
得分: 0
np.full((3,3,3), "cube")
Syntax:
numpy.full(shape, fill_value, dtype=None, order='C')
英文:
Dude, just remove ( dtype : it's optional ).
Code :
np.full( (3,3,3), "cube" )
Syntax :
numpy.full(shape, fill_value, dtype=None, order='C')
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