寻找重叠向量索引的Numpy函数

Question

看起来应该有一个numpy的函数可以用来找出两个向量的重叠部分，但我找不到。也许你们中有人知道？

这个问题用一段简单的代码来描述最清楚（见下文）。我有两组数据（x1, y1）和（x2, y2），每组的x和y都有几百个元素。我需要把它们截断，使得它们的范围相同（也就是说，x1要等于x2），y1也要对应新的x1，y2也要截断以适应新的x2。

# x1 and y1 are abscissa and ordinate from some measurement.
x1 = array([1,2,3,4,5,6,7,8,9,10])
y1 = x1**2 # I'm just making some numbers for the ordinate.

# x2 and y2 are abscissa and ordinate from a different measurement, 
# but not over the same exact range.
x2 = array([5,6,7,8,9,10,11,12,13])
y2 = sqrt(x2) # And some more numbers that aren't the same.

# And I need to do some math on just the portion where the two measurements overlap.
x3 = array([5,6,7,8,9,10])
y3 = y1[4:10] + y2[:6]

# Is there a simple function that would give me these indices, 
# or do I have to do loops and compare values?
print x1[4:10]
print x2[:6]


# ------------ THE FOLLOWING IS WHAT I WANT TO REPLACE -------------


# Doing loops is really clumsy...

# Check which vector starts lower.
if x1[0] <= x2[0]:
    # Loop through it until you find an index that matches the start of the other.
    for i in range(len(x1)):
        # Here is is.
        if x1[i] == x2[0]:
            # Note the offsets for the new starts of both vectors.
            x1off = i
            x2off = 0
            break
else:
    for i in range(len(x2)):
        if x2[i] == x1[0]:
            x1off = 0
            x2off = i
            break

# Cutoff the beginnings of the vectors as appropriate.
x1 = x1[x1off:]
y1 = y1[x1off:]
x2 = x2[x2off:]
y2 = y2[x2off:]

# Now make the lengths of the vectors be the same.
# See which is longer.
if len(x1) > len(x2):
    # Cut off the longer one to be the same length as the shorter.
    x1 = x1[:len(x2)]
    y1 = y1[:len(x2)]
elif len(x2) > len(x1):
    x2 = x2[:len(x1)]
    y2 = y2[:len(x1)]

# OK, now the domains and ranges for the two (x,y) sets are identical.    
print x1, y1
print x2, y2

谢谢！

Answer 1

对于简单的交集操作，你可以使用 np.intersect1d：

In [20]: x1 = array([1,2,3,4,5,6,7,8,9,10])

In [21]: x2 = array([5,6,7,8,9,10,11,12,13])

In [22]: x3 = np.intersect1d(x1, x2)

In [23]: x3
Out[23]: array([ 5,  6,  7,  8,  9, 10])

但看起来你需要的是不同的东西。正如 @JoranBeasley 在评论中提到的，你可以使用 np.in1d，不过你需要用两次：

这里是数据：

In [57]: x1
Out[57]: array([ 1,  2,  3,  4,  5,  6,  7,  8,  9, 10])

In [58]: y1
Out[58]: array([  1,   4,   9,  16,  25,  36,  49,  64,  81, 100])

In [59]: x2
Out[59]: array([ 5,  6,  7,  8,  9, 10, 11, 12, 13])

In [60]: y2
Out[60]: 
array([ 2.23606798,  2.44948974,  2.64575131,  2.82842712,  3.        ,
        3.16227766,  3.31662479,  3.46410162,  3.60555128])

获取 (x1, y1) 数据的子集：

In [61]: mask1 = np.in1d(x1, x2)

In [62]: xx1 = x1[mask1]

In [63]: yy1 = y1[mask1]

In [64]: xx1, yy1
Out[64]: (array([ 5,  6,  7,  8,  9, 10]), array([ 25,  36,  49,  64,  81, 100]))

获取 (x2, y2) 数据的子集。注意，这次传给 np.in1d 的参数顺序是 x2, x1：

In [65]: mask2 = np.in1d(x2, x1)

In [66]: xx2 = x2[mask2]

In [67]: yy2 = y2[mask2]

In [68]: xx2, yy2
Out[68]: 
(array([ 5,  6,  7,  8,  9, 10]),
 array([ 2.23606798,  2.44948974,  2.64575131,  2.82842712,  3.        ,
         3.16227766]))

其实我们并不需要单独形成 xx2，因为它和 xx1 是一样的。现在我们可以对 yy1 和 yy2 进行操作。例如：

In [69]: yy1 + yy2
Out[69]: 
array([  27.23606798,   38.44948974,   51.64575131,   66.82842712,
         84.        ,  103.16227766])