如何在python中对二维数组进行排序

Question

Tool = [[0 for x in xrange(3)] for y in xrange(len(xTool)-1)]
for l in xrange((len(xTool) - 1)):
  Tool[l][0] = yTool[l]; Tool[l][1] = xTool[l]; Tool[l][2] = zTool[l]  

I am starting with points coordinates, which are in 3 lists (xTool, yTool, zTool), representing respectively the x, y and z coordinates of all my points.

The aim here is to create a matrix of 3 columns and many rows (over 10,000), where each row represent the point's 3 coordinates. The next step I do is a vector transformation like so: ( This has minor importance, only if you really want to understand what i'm doing

rTool = numpy.zeros_like(Tool)   
for rt in xrange((len(xTool) - 1)):
  rTool[rt][0] = (Tool[rt][0] * cos(angle)) - (Tool[rt][1] * sin(angle))
  rTool[rt][1] = (Tool[rt][0] * sin(angle)) + (Tool[rt][1] * cos(angle))
  rTool[rt][2] = Tool[rt][2]

Finally, what I'm trying to do is order my rTool in regards to my 2nd column ([1]). For instance, I printed 5 entries of my rTool. By sorting them according to the 2nd column, the last row should be the first. I am really struggling to do this, and I suspect it's because I have tuples instead of a real 3 column mathematical matrix.

[[ -584.89837646 -3648.6472168    402.177948  ]
 [ -542.8659668  -3663.34545898   405.76959229]
 [ -500.831604   -3678.04785156   409.32122803]
 [ -458.79336548 -3692.75854492   412.7930603 ]
 [ -416.74984741 -3637.48022461   416.15090942]]

Don't hesitate to ask for clarification and I hope you will be able to help me! Thanks.

Answer 1

First and foremost, learn numpy. Doing this kind of thing in plain python goes against everything pythonic.

Once you've done that:

sorted_rtool = rTool[np.argsort(rTool[:,1])]

To drive the importance of numpy home:

rTool = np.dot(Tool, R)

Is not only a lot cleaner, its also orders of magnitude faster.

Answer 2

You could use the 'key' parameter of 'sort' or 'sorted', see https://wiki.python.org/moin/HowTo/Sorting#Key_Functions

In your case:

rTool.sort(key=lambda x: x[1])

or

rToolSorted = sorted(rTool, key=lambda x: x[1])

Answer 3

You're looking for the key keyword of list.sort() (or the builtin sorted() ) Look at the Key Functions section here .

Often people just use a lambda function for the sort key, as it's concise, and you often don't use the sort function more than once. All it is, is a function which returns an object key , on which each element of the list is sorted.

So you could do:

def keyFunc(element):
  return element[1]

rTool.sort(key=keyFunc)

or:

rTool.sort(key=lambda x: x[1])

In both cases, you could use rTool = sorted(rTool, key=...) instead, the difference being that list.sort() does an inplace sort, and is more efficient if you don't need the original array.

The key function can really be anything you want, provided it returns something else that can be sorted, so if your points were objects with x, y, z attributes, then you could do rTool.sort(key x: xy)