检查列表是否不包含列表

Question

Firstly, sorry if this is in fact a duplicate - I have spent the last three hours attempting to solve this problem and haven't been able to find any solution.

Problem

I am using lists to represent coordinates as [x, y] . I want to find out if a list of coordinates does not contain a specified coordinate. For example, if I have the list of coordinates [[3.3, 4.4], [5.5, 6.6]] and the coordinate [1.1, 2.2] , I want a return of True , because the coordinate is not in the list of coordinates.

It may be worth noting that the list of coordinates is generated using the OpenCV functions cv2.findContours() , cv2.minAreaRect() and finally cv2.boxPoints() which results in a list of lists. These coordinates are stored in a dict and accessed from there; calling a print() of the coordinates gives me the coordinates in the format [array([3.3, 4.4], dtype=float32), array([5.5, 6.6], dtype=float32)] as opposed to the format [[3.3, 4.4], [5.5, 6.6]] which is given when I print() the coordinates straight after finding them with cv2.boxPoints() .

What I have tried

I have tried to use the answer to this question but I get the error ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all() ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all() .

The code for that attempt looks like this:

for coordinate in box:
    if coordinate not in specialCoordinates:
        # operations go here

I then attempted to use the answer to this question about a.all() but I get the same error.

The code for that attempt looks like this:

for coordinate in box:
    if not all(coordinate == special for special in specialCoordinates):
        # operations go here

I also tried this:

for coordinate in box:
    if all(coordinate != special for special in specialCoordinates):
        # operations go here

Additional information

The format mentioned above if coordinate not in specialCoordinates works when I try the following in the Python 2.7 interpreter:

Python 2.7.15 (v2.7.15:ca079a3ea3, Apr 30 2018, 16:30:26) [MSC v.1500 64 bit (AMD64)] on win32

Type "help", "copyright", "credits" or "license" for more information.

>>> a = [[3.3, 4.4], [5.5, 6.6]]

>>> b = [1.1, 2.2]

>>> b not in a

True

Answer 1

The main problem here is that the numpy array elements in the list have precision associated with them for example: a = [np.array([3.3, 4.4], dtype='float32'), np.array([5.5, 6.6], dtype='float32')] Which is equivalent to: [array([ 3.29999995, 4.4000001 ], dtype=float32), array([ 5.5 , 6.5999999], dtype=float32)]

Hence, if you look for [3.3, 4.4] in a, it won't be there so you might want to play around with the precision or do a bit of casting.

Answer 2

import numpy as np
#a = [array([3.3, 4.4]), array([5.5, 6.6])]
a = np.array([i.tolist() for i in a])

def if_not_in_a(A, B):
    for i in A:
        if np.linalg.norm(i - B) == 0: 
        #you can use tolerance like < 1e-2 to avoid floating point complicacy
            return False
    return True

print(if_not_in_a(a, np.array([1.1, 2.2])))
#prints True

print(if_not_in_a(a, np.array([3.3, 4.4])))
#prints False