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) 

基本上，这会将我们的(x,y) 数据点的x 坐标存储在数组t 中；以及数组 s 中的结果 y 坐标(y=f(x) 的结果，在本例中为 sin(x))。然后，可以很方便的使用numpy.nditer函数获取t和s中连续的entry对，代表 (x,y) 数据点的坐标，如:

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

因此，我尝试将以下代码段作为 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))

...结果是:

$ 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 - 结果是 the iterator object nditer, introduced in NumPy 1.6 ，在我安装的 Python 2.7 的 numpy 版本中不可用。

因此，由于我也想支持该特定版本，因此我需要找到一种适用于较旧的 numpy 的方法 - 但我仍然希望仅指定 for x,y in somearray，并在循环中直接获取坐标。

在弄乱了 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))

...这似乎工作正常

$ 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 - 在 numpy < 1.6.0="">

最佳答案

concatenating怎么样？将两个向量放入一个数组中:

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

这给出了

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

如果你想迭代以节省内存，你可以使用 itertools.izip功能:

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

关于python - 在 numpy 中迭代两个数组，没有 nditer？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/16646456/