在numpy中迭代两个没有nditer的数组？

Consider a specification of numpy arrays, typical for specifying matplotlib plotting data:

考虑numpy数组的规范，通常用于指定matplotlib绘制数据：

t = np.arange(0.0,1.5,0.25)
s = np.sin(2*np.pi*t)

Basically, this stores the x coordinates of our (x,y) data points in the array t; and the resulting y coordinates (result of y=f(x), in this case sin(x)) in the array s. Then, it is very convenient to use the numpy.nditer function to obtain consecutive pairs of entries in t and s, representing the (x,y) coordinate of a data point, as in:

基本上，这将我们的（x，y）数据点的x坐标存储在数组t中;以及在数组s中得到的y坐标（y = f（x）的结果，在这种情况下为sin（x））。然后，使用numpy.nditer函数获取t和s中的连续条目对非常方便，表示数据点的（x，y）坐标，如下所示：

for x, y in np.nditer([t,s]):
  print("xy: %f:%f" % (x,y))

So, I'm trying the following snippet as test.py:

所以，我正在尝试以下代码片段作为test.py：

import numpy as np
print("numpy version {0}".format(np.__version__))
t = np.arange(0.0,1.5,0.25)   ; print("t", ["%+.2e"%i for i in t])
s = np.sin(2*np.pi*t)         ; print("s", ["%+.2e"%i for i in s])
print("i", ["% 9d"%i for i in range(0, len(t))])
for x, y in np.nditer([t,s]):
  print("xy: %f:%f" % (x,y))

... and the results are:

......结果如下：

$ python3.2 test.py 
numpy version 1.7.0
t ['+0.00e+00', '+2.50e-01', '+5.00e-01', '+7.50e-01', '+1.00e+00', '+1.25e+00']
s ['+0.00e+00', '+1.00e+00', '+1.22e-16', '-1.00e+00', '-2.45e-16', '+1.00e+00']
i ['        0', '        1', '        2', '        3', '        4', '        5']
xy: 0.000000:0.000000
xy: 0.250000:1.000000
xy: 0.500000:0.000000
xy: 0.750000:-1.000000
xy: 1.000000:-0.000000
xy: 1.250000:1.000000

$ python2.7 test.py 
numpy version 1.5.1
('t', ['+0.00e+00', '+2.50e-01', '+5.00e-01', '+7.50e-01', '+1.00e+00', '+1.25e+00'])
('s', ['+0.00e+00', '+1.00e+00', '+1.22e-16', '-1.00e+00', '-2.45e-16', '+1.00e+00'])
('i', ['        0', '        1', '        2', '        3', '        4', '        5'])
Traceback (most recent call last):
  File "test.py", line 10, in <module>
    for x, y in np.nditer([t,s]):
AttributeError: 'module' object has no attribute 'nditer'

Ah - it turns out, that the iterator object nditer, introduced in NumPy 1.6, is not available in the numpy version of my Python 2.7 installation.

啊 - 事实证明，在NumPy 1.6中引入的迭代器对象nditer在我的Python 2.7安装的numpy版本中不可用。

So, as I'd like to support that particular version too, I'd need to find a way working for older numpy - but I'd still like the convenience of just specifying for x,y in somearray, and getting the coordinates directly in the loop.

所以，因为我也想支持那个特定的版本，我需要找到一种适用于较旧的numpy的方法 - 但我仍然喜欢在somearray中指定x，y的便利，并直接获取坐标在循环。

After some messing about with numpy documentation, I came up with this getXyIter function:

在讨论了numpy文档之后，我想出了这个getXyIter函数：

import numpy as np
print("numpy version {0}".format(np.__version__))
t = np.arange(0.0,1.5,0.25)   ; print("t", ["%+.2e"%i for i in t])
s = np.sin(2*np.pi*t)         ; print("s", ["%+.2e"%i for i in s])
print("i", ["% 9d"%i for i in range(0, len(t))])

def getXyIter(inarr):
  if np.__version__ >= "1.6.0":
    return np.nditer(inarr.tolist())
  else:
    dimensions = inarr.shape
    xlen = dimensions[1]
    xinds = np.arange(0, xlen, 1)
    return np.transpose(np.take(inarr, xinds, axis=1))

for x, y in getXyIter(np.array([t,s])):
  print("xyIt: %f:%f" % (x,y))

for x, y in np.nditer([t,s]):
  print("xynd: %f:%f" % (x,y))

... which seems to work fine

......似乎工作正常

$ python2.7 test.py 
numpy version 1.5.1
('t', ['+0.00e+00', '+2.50e-01', '+5.00e-01', '+7.50e-01', '+1.00e+00', '+1.25e+00'])
('s', ['+0.00e+00', '+1.00e+00', '+1.22e-16', '-1.00e+00', '-2.45e-16', '+1.00e+00'])
('i', ['        0', '        1', '        2', '        3', '        4', '        5'])
xyIt: 0.000000:0.000000
xyIt: 0.250000:1.000000
xyIt: 0.500000:0.000000
xyIt: 0.750000:-1.000000
xyIt: 1.000000:-0.000000
xyIt: 1.250000:1.000000
Traceback (most recent call last):
  File "test.py", line 23, in <module>
    for x, y in np.nditer([t,s]):
AttributeError: 'module' object has no attribute 'nditer'
$ python3.2 test.py 
numpy version 1.7.0
t ['+0.00e+00', '+2.50e-01', '+5.00e-01', '+7.50e-01', '+1.00e+00', '+1.25e+00']
s ['+0.00e+00', '+1.00e+00', '+1.22e-16', '-1.00e+00', '-2.45e-16', '+1.00e+00']
i ['        0', '        1', '        2', '        3', '        4', '        5']
xyIt: 0.000000:0.000000
xyIt: 0.250000:1.000000
xyIt: 0.500000:0.000000
xyIt: 0.750000:-1.000000
xyIt: 1.000000:-0.000000
xyIt: 1.250000:1.000000
xynd: 0.000000:0.000000
xynd: 0.250000:1.000000
xynd: 0.500000:0.000000
xynd: 0.750000:-1.000000
xynd: 1.000000:-0.000000
xynd: 1.250000:1.000000

My question is - is this the way, this kind of iteration is supposed to be done, in versions of numpy < 1.6.0?

我的问题是 - 就这样，这种迭代应该在numpy <1.6.0的版本中完成吗？

2 个解决方案

#1

How about concatenating the two vectors into an array:

如何将两个向量连接成一个数组：

for x,y in np.c_[t,s]:
    print("xy: %f:%f" % (x,y))

This gives

这给了

xy: 0.000000:0.000000
xy: 0.250000:1.000000
xy: 0.500000:0.000000
xy: 0.750000:-1.000000
xy: 1.000000:-0.000000
xy: 1.250000:1.000000

If you want to iterate so you can save memory, you can use the itertools.izip function:

如果要迭代以便节省内存，可以使用itertools.izip函数：

for x,y in itertools.izip(t,s):
    print("xy: %f:%f" % (x,y))

#2

for x, y in zip(t,s):. For 1d arrays, it's really that simple.

for x，y in zip（t，s）:.对于1d阵列，它真的很简单。

Verified to work in both Python 2 and Python 3. zip() returns a list on Python2 though, so as DiggyF suggests, itertools.izip() may be more appropriate for large arrays.

已验证可在Python 2和Python 3中工作.zip（）返回Python2上的列表，因此DiggyF建议，itertools.izip（）可能更适合大型数组。

For >1D arrays, iteration moves through the last dimension returning (N-1)D arrays. If you have to deal with N-d arrays, this may or may not be what you want.

对于> 1D阵列，迭代移动通过返回（N-1）D阵列的最后一维。如果你必须处理N-d阵列，这可能是也可能不是你想要的。

Regardless, it's unquestionably portable, and iterating on array objects is intended to support this kind of usecase :)

无论如何，它毫无疑问是可移植的，并且迭代数组对象旨在支持这种用例:)

#1