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知识问答

如何在python中识别numpy类型？

问题：如何在python中识别numpy类型？

如何可靠地确定一个对象是否具有numpy类型？

我意识到这个问题与鸭子类型的哲学背道而驰，但是我们的想法是确保一个函数（使用scipy和numpy）永远不会返回一个numpy类型，除非使用numpy类型进行调用。 这是另一个问题的解决方案，但是我认为确定对象是否具有numpy类型的一般问题与原始问题相距甚远，因此应将它们分开。

点击查看英文原文

How can one reliably determine if an object has a numpy type?

I realize that this question goes against the philosophy of duck typing, but idea is to make sure a function (which uses scipy and numpy) never returns a numpy type unless it is called with a numpy type. This comes up in the solution to another question, but I think the general problem of determining if an object has a numpy type is far enough away from that original question that they should be separated.

回答 0

使用内置type函数获取类型，然后可以使用该__module__属性找出定义的位置：

>>> import numpy as np
a = np.array([1, 2, 3])
>>> type(a)
<type 'numpy.ndarray'>
>>> type(a).__module__
'numpy'
>>> type(a).__module__ == np.__name__
True
点击查看英文原文

Use the builtin type function to get the type, then you can use the __module__ property to find out where it was defined:

>>> import numpy as np
a = np.array([1, 2, 3])
>>> type(a)
<type 'numpy.ndarray'>
>>> type(a).__module__
'numpy'
>>> type(a).__module__ == np.__name__
True

回答 1

我想出的解决方案是：

isinstance(y, (np.ndarray, np.generic) )

但是，并不是所有的numpy类型都保证是np.ndarray或都100％清晰，np.generic这可能不是版本可靠的。

点击查看英文原文

The solution I’ve come up with is:

isinstance(y, (np.ndarray, np.generic) )

However, it’s not 100% clear that all numpy types are guaranteed to be either np.ndarray or np.generic, and this probably isn’t version robust.

回答 2

老问题，但我想出一个明确的答案，举一个例子。我也遇到了同样的问题，也没有找到明确的答案，因此可以使问题保持​​新鲜感。关键是确保已numpy导入，然后运行isinstance布尔。虽然这看起来很简单，但是如果您要对不同的数据类型进行一些计算，则在开始一些numpy向量化操作之前，此小检查可以作为一项快速测试。

##################
# important part!
##################

import numpy as np

####################
# toy array for demo
####################

arr = np.asarray(range(1,100,2))

########################
# The instance check
######################## 

isinstance(arr,np.ndarray)
点击查看英文原文

Old question but I came up with a definitive answer with an example. Can’t hurt to keep questions fresh as I had this same problem and didn’t find a clear answer. The key is to make sure you have numpy imported, and then run the isinstance bool. While this may seem simple, if you are doing some computations across different data types, this small check can serve as a quick test before your start some numpy vectorized operation. 

##################
# important part!
##################

import numpy as np

####################
# toy array for demo
####################

arr = np.asarray(range(1,100,2))

########################
# The instance check
######################## 

isinstance(arr,np.ndarray)

回答 3

这实际上取决于您要查找的内容。

  • 如果要测试序列是否实际上是a ndarray，则a isinstance(..., np.ndarray)可能是最简单的。确保您不要在后台重新加载numpy，因为模块可能有所不同，但否则应该没问题。MaskedArrays，matrix，recarray是所有子类ndarray，所以你应该设置。
  • 如果要测试标量是否为numpy标量，事情会变得更加复杂。您可以检查它是否具有shape和dtype属性。您可以将其dtype与基本dtype 进行比较，可以在中找到其基本列表np.core.numerictypes.genericTypeRank。请注意，此列表的元素是字符串，因此您必须执行tested.dtype is np.dtype(an_element_of_the_list)…
点击查看英文原文

That actually depends on what you’re looking for.

  • If you want to test whether a sequence is actually a ndarray, a isinstance(..., np.ndarray) is probably the easiest. Make sure you don’t reload numpy in the background as the module may be different, but otherwise, you should be OK. MaskedArrays, matrix, recarray are all subclasses of ndarray, so you should be set.
  • If you want to test whether a scalar is a numpy scalar, things get a bit more complicated. You could check whether it has a shape and a dtype attribute. You can compare its dtype to the basic dtypes, whose list you can find in np.core.numerictypes.genericTypeRank. Note that the elements of this list are strings, so you’d have to do a tested.dtype is np.dtype(an_element_of_the_list)…

回答 4

要获取类型，请使用内置type函数。使用in运算符，可以检查类型是否为numpy类型，方法是检查其是否包含字符串numpy;

In [1]: import numpy as np

In [2]: a = np.array([1, 2, 3])

In [3]: type(a)
Out[3]: <type 'numpy.ndarray'>

In [4]: 'numpy' in str(type(a))
Out[4]: True

（顺便说一下，此示例在IPython中运行。非常便于交互使用和快速测试。）

点击查看英文原文

To get the type, use the builtin type function. With the in operator, you can test if the type is a numpy type by checking if it contains the string numpy;

In [1]: import numpy as np

In [2]: a = np.array([1, 2, 3])

In [3]: type(a)
Out[3]: <type 'numpy.ndarray'>

In [4]: 'numpy' in str(type(a))
Out[4]: True

(This example was run in IPython, by the way. Very handy for interactive use and quick tests.)

回答 5

请注意，它本身type(numpy.ndarray)就是一个type，请注意布尔和标量类型。如果不是直观或简单的方法，不要太气disc，起初是很痛苦的。

另请参阅：-https : //docs.scipy.org/doc/numpy-1.15.1/reference/arrays.dtypes.html-https : //github.com/machinalis/mypy-data/tree/master/numpy- py

>>> import numpy as np
>>> np.ndarray
<class 'numpy.ndarray'>
>>> type(np.ndarray)
<class 'type'>
>>> a = np.linspace(1,25)
>>> type(a)
<class 'numpy.ndarray'>
>>> type(a) == type(np.ndarray)
False
>>> type(a) == np.ndarray
True
>>> isinstance(a, np.ndarray)
True

布尔值的乐趣：

>>> b = a.astype('int32') == 11
>>> b[0]
False
>>> isinstance(b[0], bool)
False
>>> isinstance(b[0], np.bool)
False
>>> isinstance(b[0], np.bool_)
True
>>> isinstance(b[0], np.bool8)
True
>>> b[0].dtype == np.bool
True
>>> b[0].dtype == bool  # python equivalent
True

标量类型的更多乐趣，请参见：-https : //docs.scipy.org/doc/numpy-1.15.1/reference/arrays.scalars.html#arrays-scalars-built-in

>>> x = np.array([1,], dtype=np.uint64)
>>> x[0].dtype
dtype('uint64')
>>> isinstance(x[0], np.uint64)
True
>>> isinstance(x[0], np.integer)
True  # generic integer
>>> isinstance(x[0], int)
False  # but not a python int in this case

# Try matching the `kind` strings, e.g.
>>> np.dtype('bool').kind                                                                                           
'b'
>>> np.dtype('int64').kind                                                                                          
'i'
>>> np.dtype('float').kind                                                                                          
'f'
>>> np.dtype('half').kind                                                                                           
'f'

# But be weary of matching dtypes
>>> np.integer
<class 'numpy.integer'>
>>> np.dtype(np.integer)
dtype('int64')
>>> x[0].dtype == np.dtype(np.integer)
False

# Down these paths there be dragons:

# the .dtype attribute returns a kind of dtype, not a specific dtype
>>> isinstance(x[0].dtype, np.dtype)
True
>>> isinstance(x[0].dtype, np.uint64)
False  
>>> isinstance(x[0].dtype, np.dtype(np.uint64))
Traceback (most recent call last):
  File "<console>", line 1, in <module>
TypeError: isinstance() arg 2 must be a type or tuple of types
# yea, don't go there
>>> isinstance(x[0].dtype, np.int_)
False  # again, confusing the .dtype with a specific dtype


# Inequalities can be tricky, although they might
# work sometimes, try to avoid these idioms:

>>> x[0].dtype <= np.dtype(np.uint64)
True
>>> x[0].dtype <= np.dtype(np.float)
True
>>> x[0].dtype <= np.dtype(np.half)
False  # just when things were going well
>>> x[0].dtype <= np.dtype(np.float16)
False  # oh boy
>>> x[0].dtype == np.int
False  # ya, no luck here either
>>> x[0].dtype == np.int_
False  # or here
>>> x[0].dtype == np.uint64
True  # have to end on a good note!
点击查看英文原文

Note that the type(numpy.ndarray) is a type itself and watch out for boolean and scalar types. Don’t be too discouraged if it’s not intuitive or easy, it’s a pain at first.

See also: – https://docs.scipy.org/doc/numpy-1.15.1/reference/arrays.dtypes.html – https://github.com/machinalis/mypy-data/tree/master/numpy-mypy

>>> import numpy as np
>>> np.ndarray
<class 'numpy.ndarray'>
>>> type(np.ndarray)
<class 'type'>
>>> a = np.linspace(1,25)
>>> type(a)
<class 'numpy.ndarray'>
>>> type(a) == type(np.ndarray)
False
>>> type(a) == np.ndarray
True
>>> isinstance(a, np.ndarray)
True

Fun with booleans:

>>> b = a.astype('int32') == 11
>>> b[0]
False
>>> isinstance(b[0], bool)
False
>>> isinstance(b[0], np.bool)
False
>>> isinstance(b[0], np.bool_)
True
>>> isinstance(b[0], np.bool8)
True
>>> b[0].dtype == np.bool
True
>>> b[0].dtype == bool  # python equivalent
True

More fun with scalar types, see: – https://docs.scipy.org/doc/numpy-1.15.1/reference/arrays.scalars.html#arrays-scalars-built-in

>>> x = np.array([1,], dtype=np.uint64)
>>> x[0].dtype
dtype('uint64')
>>> isinstance(x[0], np.uint64)
True
>>> isinstance(x[0], np.integer)
True  # generic integer
>>> isinstance(x[0], int)
False  # but not a python int in this case

# Try matching the `kind` strings, e.g.
>>> np.dtype('bool').kind                                                                                           
'b'
>>> np.dtype('int64').kind                                                                                          
'i'
>>> np.dtype('float').kind                                                                                          
'f'
>>> np.dtype('half').kind                                                                                           
'f'

# But be weary of matching dtypes
>>> np.integer
<class 'numpy.integer'>
>>> np.dtype(np.integer)
dtype('int64')
>>> x[0].dtype == np.dtype(np.integer)
False

# Down these paths there be dragons:

# the .dtype attribute returns a kind of dtype, not a specific dtype
>>> isinstance(x[0].dtype, np.dtype)
True
>>> isinstance(x[0].dtype, np.uint64)
False  
>>> isinstance(x[0].dtype, np.dtype(np.uint64))
Traceback (most recent call last):
  File "<console>", line 1, in <module>
TypeError: isinstance() arg 2 must be a type or tuple of types
# yea, don't go there
>>> isinstance(x[0].dtype, np.int_)
False  # again, confusing the .dtype with a specific dtype


# Inequalities can be tricky, although they might
# work sometimes, try to avoid these idioms:

>>> x[0].dtype <= np.dtype(np.uint64)
True
>>> x[0].dtype <= np.dtype(np.float)
True
>>> x[0].dtype <= np.dtype(np.half)
False  # just when things were going well
>>> x[0].dtype <= np.dtype(np.float16)
False  # oh boy
>>> x[0].dtype == np.int
False  # ya, no luck here either
>>> x[0].dtype == np.int_
False  # or here
>>> x[0].dtype == np.uint64
True  # have to end on a good note!
知识问答

将Python argparse.Namespace（）视为字典的正确方法是什么？

问题：将Python argparse.Namespace（）视为字典的正确方法是什么？

如果我想将的结果（argparse.ArgumentParser()即Namespace对象）与需要字典或类映射对象的方法一起使用（请参阅collections.Mapping），那么正确的方法是什么？

C:\>python
Python 2.7.3 (default, Apr 10 2012, 23:31:26) [MSC v.1500 32 bit (Intel)] on win
32
Type "help", "copyright", "credits" or "license" for more information.
>>> import argparse
>>> args = argparse.Namespace()
>>> args.foo = 1
>>> args.bar = [1,2,3]
>>> args.baz = 'yippee'
>>> args['baz']
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: 'Namespace' object has no attribute '__getitem__'
>>> dir(args)
['__class__', '__contains__', '__delattr__', '__dict__', '__doc__', '__eq__', '_
_format__', '__getattribute__', '__hash__', '__init__', '__module__', '__ne__',
'__new__', '__reduce__', '__reduce_ex__', '__repr__', '__setattr__', '__sizeof__
', '__str__', '__subclasshook__', '__weakref__', '_get_args', '_get_kwargs', 'ba
r', 'baz', 'foo']

“进入”对象并使用其__dict__属性是否合适？

我认为答案是否定的：__dict__闻起来像实现的约定，而不是接口的含义__getattribute__，__setattr__或者说__contains__是。

点击查看英文原文

If I want to use the results of argparse.ArgumentParser(), which is a Namespace object, with a method that expects a dictionary or mapping-like object (see collections.Mapping), what is the right way to do it?

C:\>python
Python 2.7.3 (default, Apr 10 2012, 23:31:26) [MSC v.1500 32 bit (Intel)] on win
32
Type "help", "copyright", "credits" or "license" for more information.
>>> import argparse
>>> args = argparse.Namespace()
>>> args.foo = 1
>>> args.bar = [1,2,3]
>>> args.baz = 'yippee'
>>> args['baz']
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: 'Namespace' object has no attribute '__getitem__'
>>> dir(args)
['__class__', '__contains__', '__delattr__', '__dict__', '__doc__', '__eq__', '_
_format__', '__getattribute__', '__hash__', '__init__', '__module__', '__ne__',
'__new__', '__reduce__', '__reduce_ex__', '__repr__', '__setattr__', '__sizeof__
', '__str__', '__subclasshook__', '__weakref__', '_get_args', '_get_kwargs', 'ba
r', 'baz', 'foo']

Is it proper to “reach into” an object and use its __dict__ property?

I would think the answer is no: __dict__ smells like a convention for implementation, but not for an interface, the way __getattribute__ or __setattr__ or __contains__ seem to be.

回答 0

您可以使用vars（）访问命名空间的字典：

>>> import argparse
>>> args = argparse.Namespace()
>>> args.foo = 1
>>> args.bar = [1,2,3]
>>> d = vars(args)
>>> d
{'foo': 1, 'bar': [1, 2, 3]}

您可以根据需要直接修改字典：

>>> d['baz'] = 'store me'
>>> args.baz
'store me'

是的，可以访问__dict__属性。这是定义明确，经过测试且有保证的行为。

点击查看英文原文

You can access the namespace’s dictionary with vars():

>>> import argparse
>>> args = argparse.Namespace()
>>> args.foo = 1
>>> args.bar = [1,2,3]
>>> d = vars(args)
>>> d
{'foo': 1, 'bar': [1, 2, 3]}

You can modify the dictionary directly if you wish:

>>> d['baz'] = 'store me'
>>> args.baz
'store me'

Yes, it is okay to access the __dict__ attribute. It is a well-defined, tested, and guaranteed behavior.

回答 1

从马口直行：

如果您更喜欢具有类似dict的属性视图，则可以使用标准的Python习惯用法vars()：

>>> parser = argparse.ArgumentParser()
>>> parser.add_argument('--foo')
>>> args = parser.parse_args(['--foo', 'BAR'])
>>> vars(args)
{'foo': 'BAR'}

— Python标准库，16.4.4.6。命名空间对象

点击查看英文原文

Straight from the horse’s mouth:

If you prefer to have dict-like view of the attributes, you can use the standard Python idiom, vars():

>>> parser = argparse.ArgumentParser()
>>> parser.add_argument('--foo')
>>> args = parser.parse_args(['--foo', 'BAR'])
>>> vars(args)
{'foo': 'BAR'}

— The Python Standard Library, 16.4.4.6. The Namespace object

回答 2

“进入”对象并使用其dict属性是否合适？

通常，我会说“不”。但是Namespace，令我震惊的是过度设计，可能是因为类无法从内置类型继承。

另一方面，Namespace确实提供了一种针对argparse的面向任务的方法，我想不出需要抓住的情况__dict__，但是我的想象力的极限与您的一样。

点击查看英文原文

Is it proper to “reach into” an object and use its dict property?

In general, I would say “no”. However Namespace has struck me as over-engineered, possibly from when classes couldn’t inherit from built-in types.

On the other hand, Namespace does present a task-oriented approach to argparse, and I can’t think of a situation that would call for grabbing the __dict__, but the limits of my imagination are not the same as yours.