在python 3中将map对象转换为numpy数组
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/28524378/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Convert map object to numpy array in python 3
提问by btel
In Python 2 I could do the following:
在 Python 2 中，我可以执行以下操作：
import numpy as np
f = lambda x: x**2
seq = map(f, xrange(5))
seq = np.array(seq)
print seq
# prints: [ 0 1 4 9 16]
In Python 3 it does not work anymore:
在 Python 3 中它不再起作用：
import numpy as np
f = lambda x: x**2
seq = map(f, range(5))
seq = np.array(seq)
print(seq)
# prints: <map object at 0x10341e310>
How do I get the old behaviour (converting the mapresults to numpyarray)?
如何获得旧行为（将map结果转换为numpy数组）？
Edit: As @jonrsharpe pointed out in his answer this could be fixed if I converted seqto a list first:
编辑：正如@jonrsharpe 在他的回答中指出的那样，如果我先转换seq为列表，这可以解决：
seq = np.array(list(seq))
but I would prefer to avoid the extra call to list.
但我更愿意避免额外调用list.
回答by jonrsharpe
Although you refer to it as seq, the mapobject in Python 3 is nota sequence(it's an iterator, see what's new in Python 3). numpy.arrayneeds a sequence so the lencan be determined and the appropriate amount of memory reserved; it won't consume an iterator. For example, the rangeobject, which doessupport most sequence operations, canbe passed directly;
虽然你把它称作seq时，map在Python 3的对象是不是一个序列（这是一个迭代器，看看有什么在Python 3的新功能）。numpy.array需要一个序列，以便len可以确定并保留适当的内存量；它不会消耗迭代器。例如，range对象，它不支持大多数序列的操作，可以被直接传递;
seq = np.array(range(5))
print(seq)
# prints: [0 1 2 3 4]
To restore the previous behaviour, as you're aware, you can explicitly convert the mapobject back to a sequence (e.g. list or tuple):
如您所知，要恢复以前的行为，您可以将map对象显式转换回序列（例如列表或元组）：
seq = np.array(list(seq)) # should probably change the name!
However, as the documentationputs it:
但是，正如文档所说：
a quick fix is to wrap
map()inlist(), e.g.list(map(...)), but a better fix is often to use a list comprehension (especially when the original code useslambda)
速战速决是包装
map()中list()，如list(map(...))，但更好的解决办法是经常使用列表理解（尤其是当原来的代码使用lambda）
So another option would be:
所以另一种选择是：
seq = [f(x) for x in range(5)]
or just:
要不就：
seq = [x**2 for x in range(5)]
Alternatively, actually use numpyfrom the start:
或者，numpy从一开始就实际使用：
import numpy as np
arr = np.arange(5)
arr **= 2
print(arr)
# prints [ 0 1 4 9 16] in 2.x and 3.x
回答by Oliver W.
One more alternative, other than the valid solutions @jonrsharpe already pointed out is to use np.fromiter:
除了@jonrsharpe 已经指出的有效解决方案之外，另一种选择是使用np.fromiter：
>>> import numpy as np
>>> f = lambda x: x**2
>>> seq = map(f, range(5))
>>> np.fromiter(seq, dtype=np.int)
array([ 0, 1, 4, 9, 16])
