any（）python中的多个for语句

Question

So, I have four lists. Two hold x coordinates ( foodX and newPosXArray ) and the other two hold y coordinates ( foodX and newPosYArray ). Both the food and the newPos arrays are of different dimensions because I have multiple "food sources", and multiple objects searching for the food. I want to write an if statement that does something when the objects get to within a certain distance of the food. My attempt with any()

if any(np.sqrt((newPosXArray[u]-foodX[t])**2 + (newPosYArray[u]-foodY[t])**2) <= 0.2 for t in zip(food[0], food[1]) for u in zip(newPosXArray, newPosYArray)):
#dosomething

I am getting an error TypeError: list indices must be integers, not tuple

Edit: maybe I am misunderstanding zip(). I was assuming that it condenses this

 if any(np.sqrt((newPosXArray[u]-foodX[t])**2 + (newPosYArray[u]-foodY[t])**2) <= 0.2 for t in foodX for t in foodY for u in newPosXArray for u in newPosYArray):

Typical Values of what I am working with

foodX = [5,5,-5,-5]
foodY = [5,-5,5,-5]

In [65]: newPosXArray
Out[65]:
[-0.012880860643600167,
 -0.566815786730638,
 0.7905336304903405,
 0.09006991095474826,
 0.26518652615441063,
 0.3161232055076695,
 0.681255361368023,
 -0.6849985596071202,
 0.7140832628874829,
 0.4958515031060564]


In [66]: newPosYArray
Out[66]: 
[-0.41112817779983235,
 -0.08554837651693648,
 0.8743935617169996,
 -0.9384733737088207,
 0.02423386678116546,
 -0.3735855691077572,
 -0.5251118585489394,
 0.3950871276165102,
 0.9892320167752822,
 -0.7342372054958279]

of course, none of these values will return true in the if statement because none are within a 0.2 radius of any of the food coordinates

Answer 1

您尝试获取食物[] [z]，但是z是元组（通过zip方法），也许您打算这样做

food[0][z[0]] 

Answer 2

The problem is that u and t are tuples because zip places x and y side by side in tuple format. You need to say food[0][t[1]] and food[1][t[0]] where you intend to use the list values as coordinates.

From what it sees, you are trying to give it an x,(x,y) coordinate instead of an x,y coordinate.

This holds where you are using u as a list index as well.

if any(np.sqrt((u[0] - t[0]) ** 2 + (u[1] - t[1]) ** 2) <= 0.2 for t in zip(foodX,foodY) for u in zip(newPosXArray,newPosYArray))

From values given it evaluates to False until the 0.2 value gets to 5, which sounds weird given how low your initial value was, but that's how the distance formula works.

Alternatively for readability you could change u to new and t to current .

Answer 3

Just to be clear nested loops and zip do not do the same thing:

>>> [(i, j) for i in range(3) for j in range(3)]
[(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)]

Whereas:

>>> list(zip(range(3), range(3)))
[(0, 0), (1, 1), (2, 2)]

itertools.product does the same as next for loops:

>>> import itertools
>>> list(itertools.product(range(3), range(3)))
[(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)]

You don't index into the array when you iterate over an array, eg:

>>> food = [1,2,3,4]
>>> [f+1 for f in food]     # Note you did [food[f] for f in food]
[2,3,4,5]

So fixing your second example looks like:

if any(np.sqrt((u_x-t_x)**2 + (u_y-t_y)**2) <= 0.2 
       for t_x in foodX for t_y in foodY for u_x in newPosXArray for u_y in newPosYArray):

But this iterates foodY for every foodX so assume you do want these to be zipped which would be written:

if any(np.sqrt((u_x-t_x)**2 + (u_y-t_y)**2) <= 0.2 
       for t_x, t_y in zip(foodX, foodY) for u_x, u_y in zip(newPosXArray, newPosYArray)):

Or using itertools.product (which personally I don't find any easier to follow in this instance:

import itertools
if any(np.sqrt((u_x-t_x)**2 + (u_y-t_y)**2) <= 0.2
       for (t_x, t_y), (u_x, u_y) in itertools.product(zip(foodX, foodY), zip(newPosXArray, newPosYArray))):

Answer 4

Of course, Pythonic way is to just use complex numbers instead of those 2D data separated along the wrong axis. But if you really need to use that data format, I guess

any(math.hypot(X-x,Y-y)<.2 for X,Y in zip(foodX,foodY) for x,y in zip(posX,posY))

will do what you want.