在numpy中迭代两个没有nditer的数组？

Question

考虑一个numpy数组规范,通常用于指定matplotlib绘图数据:

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

基本上,这会将数据点的x坐标存储(x,y)在数组中t; 并且在数组中得到的y坐标(在这种情况下为y = f(x)的结果sin(x))s.然后,使用该numpy.nditer函数获取连续的条目对t并且s表示(x,y)数据点的坐标非常方便,如:

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

所以,我正在尝试以下代码段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))

......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'

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

所以,由于我也想支持那个特定的版本,我需要找到适合老版本的方法numpy- 但我仍然喜欢只是指定的方便for x,y in somearray,并直接在循环中获取坐标.

在对numpy文档进行一些搞乱之后,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))

......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

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？

Answer 1

如何将两个向量串联成一个数组：

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

这给

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

如果要迭代以节省内存，可以使用以下itertools.izip函数：

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