找到几个相同的两个数组的元组成员？

Question

I have 2 lists of positional tuples (x,y data). I would like to return 2 arrays or lists of the indexes for positions (or a tuple) that are in both lists. However the positional data values will not be exactly equal, there will be an uncertianty of +/- 4 on both the x and y coordinates.

For example:

A=[(1168.593,9.874), (1799.244,40.201),(780.533,12.636)]
B=[(1170.909,8.194), (793.149,10.885), (1801.493,41.603)]

it should return:

c=[(0,0),(1,2)]

or:

d=[0,1] #indexes for A
e=[0,2] #indexes for B

Either one would be fine to use.

Is there a function in Python you can use that returns the indexes of matching data in 2 lists, by specifying a +/- value as well?

I need to do this for 3 lists of ~400 tuples each, which are not equal in size.

I was thinking of even using something like:

common=[a in A for a in B]

and somehow specifying a range for a , just looking at the x and y data and return the indices instead of true/false, but I really don't know how to approach this. Is a loop the only way to do this, by looking at each value separately, getting a difference between them and seeing if this is < 4, then getting the indexes?

Answer 1

也许你可以考虑使用math模块中的函数isclose ？

Answer 2

How about a brute-force solution?

In [5]: c = []

In [6]: for i, (x1, y1) in enumerate(A):
   ...:     for j, (x2, y2) in enumerate(B):
   ...:         if (x1 - 4 <= x2 <= x1 + 4) and (y1 - 4 <= y2 <= y1 + 4):
   ...:             c.append((i,j))
   ...:
   ...:

In [7]: c
Out[7]: [(0, 0), (1, 2)]

Of course, you can replace the conditional with whatever you want. Likely, using math.isclose is a good idea. There may be a better numpy way of doing this in a vectorized fashion. But this should work if efficiency isn't a concern.