Question

当我使用map和lambda函数时，我仍然坚持如何遍历配对列表。我想根据中心位置和所选位置（x，y）到中心的距离以及特定距离出现的次数创建一系列直方图，但我不断得到索引超出范围错误而且我不知道不明白为什么。我不知道如何迭代我需要指定哪两个值的位置。整个事情除n部分外都有效。

很抱歉没有更清楚，locations = numpy.array（（x，y））是一个布尔数组的位置，它产生我想要测试的特定位置，而不是整个数组。产生的值（x，y）是两行数组，其中我想要的值按列配对。之前的代码是：

     def detect_peaks(data):
        average=numpy.average(data)*2
        local_max = data > average
        return local_max

    (x,y) = numpy.where(detect_peaks(data))

    for test_x in range(0, 8):
       for test_y in range(0,8):
          distances=[]
          locations=numpy.array((x,y))
          central=numpy.array((test_x,test_y))
          [map(lambda x1: distances.append(numpy.linalg.norm(locations[(x1,0),  (x1,1)]-central)), n) for n in locations]
          histogram=numpy.histogram(distances, bins=10)

我将重写map / lambda函数并返回。谢谢！

Answer 1

这是你想要的吗？ x和y是int的数组，而不是float。不是我喜欢双for循环，它们应该用更多的矢量化算法替换，但为了保持变化最小化和高亮度有问题的行，这里就是：

>>> a2
array([[ 0.92607265,  1.26155686,  0.31516174,  0.91750943,  0.30713193],
       [ 1.0601752 ,  1.10404664,  0.67766044,  0.36434503,  0.64966887],
       [ 1.29878971,  0.66417322,  0.48084284,  1.0956822 ,  0.27142741],
       [ 0.27654032,  0.29566566,  0.33565457,  0.29749312,  0.34113315],
       [ 0.33608323,  0.25230828,  0.41507646,  0.37872512,  0.60471438]])
>>> numpy.where(detect_peaks(a2))
(array([0, 2]), array([1, 0]))
>>> def func1(locations): #You don't need to unpack the numpy.where result.
    for test_x in range(0, 4):
       for test_y in range(0, 4):
          locations=numpy.array(locations)
          central=numpy.array((test_x,test_y))
          #Vectorization is almost always better.
          #Be careful, iterate an array means iterate its rows, so, transpose it first.
          distances=map(numpy.linalg.norm, (locations-central.reshape((2,-1))).T)
          histogram=numpy.histogram(distances, bins=10)
          print 'cental point:', central
          print 'distance lst:', distances
          print histogram
          print '-------------------------'

结果：

>>> func1(numpy.where(detect_peaks(a2)))
cental point: [0 0]
distance lst: [1.0, 2.0]
(array([1, 0, 0, 0, 0, 0, 0, 0, 0, 1]), array([ 1. ,  1.1,  1.2,  1.3,  1.4,  1.5,  1.6,  1.7, 1.8,  1.9,  2. ]))
-------------------------
cental point: [0 1]
distance lst: [0.0, 2.2360679774997898]
(array([1, 0, 0, 0, 0, 0, 0, 0, 0, 1]), array([ 0.        ,  0.2236068 ,  0.4472136 ,  0.67082039,  0.89442719, 1.11803399,  1.34164079,  1.56524758,  1.78885438,  2.01246118,         2.23606798]))
-------------------------#and more

Answer 2

想出了这个：

def detect_peaks(arrayfinal):
    average=numpy.average(arrayfinal)
    local_max = arrayfinal > average
    return local_max

def dist(distances, center, n):
   distance=numpy.linalg.norm(n-center)
   distances.append(distance)

def histotest():
   peaks = numpy.where(detect_peaks(arrayfinal))
   ordered= zip(peaks[0],peaks[1])
   for test_x in range(0, 2048):
        for test_y in range(0,2048):
            distances=[]
            center=numpy.array((test_x,test_y))
            [dist(distances, center, n) for n in ordered]
            histogram=numpy.histogram(distances, bins=30)
            print histogram

它似乎有用，但我更喜欢你的。

使用map和lambda时如何遍历配对列表？

2 个答案: