如何仅将字典中的值选择/格式化为列表或numpy数组? [英] How I select/format only the values from a dictionary into a list or numpy array?

问题描述

如何获取仅打印平均值的列表? 我只需要它与我的np完全相同的格式 数组，以便我可以比较它们是否相同.

How do I get it to print just a list of the averages? I just need it to be the exact same format as my np arrays so I can compare them to see if they are the same or not.

代码:

import numpy as np
from pprint import pprint

centroids = np.array([[3,44],[4,15],[5,15]])
dataPoints = np.array([[2,4],[17,4],[45,2],[45,7],[16,32],[32,14],[20,56],[68,33]])

def size(vector):
    return np.sqrt(sum(x**2 for x in vector))

def distance(vector1, vector2):
    return size(vector1 - vector2)

def distances(array1, array2):
    lists = [[distance(vector1, vector2) for vector2 in array2] for vector1 in array1]
    #print lists.index(min, zip(*lists))
    smallest = [min(zip(l,range(len(l)))) for l in zip(*lists)]
    clusters = {}
    for j, (_, i) in enumerate(smallest):
        clusters.setdefault(i,[]).append(dataPoints[j])
    pprint (clusters)
    print'\nAverage of Each Point'
    avgDict = {}
    for k,v in clusters.iteritems():
        avgDict[k] = sum(v)/ (len(v))
    avgList = np.asarray(avgDict)
    pprint (avgList)

distances(centroids,dataPoints)

当前输出:

{0: [array([16, 32]), array([20, 56])],
 1: [array([2, 4])],
 2: [array([17,  4]),
     array([45,  2]),
     array([45,  7]),
     array([32, 14]),
     array([68, 33])]}

Average of Each Point
array({0: array([18, 44]), 1: array([2, 4]), 2: array([41, 12])}, dtype=object)

所需的输出:

[[18,44],[2,4],[41,12]]

或者是比较我的阵列/列表的最佳格式.我知道我应该只使用一种数据类型.

Or whatever the best format to compare my arrays/lists. I am aware I should have just stuck with one data type.

推荐答案

您是否尝试通过最接近的centroids的索引对dataPoints进行聚类，并找出聚类点的平均位置?如果是这样，我建议您使用一些numpy广播规则来获取所需的输出.

Do you try to cluster the dataPoints by the index of the nearest centroids, and find out the average position of the clustered points? If it is, I advise to use some broadcast rules of numpy to get the output you need.

考虑这个

    np.linalg.norm(centroids[None, :, :] - dataPoints[:, None, :], axis=-1)

它创建一个矩阵，显示dataPoints和centroids之间的所有距离，

It creates a matrix showing all distances between dataPoints and centroids,

    array([[ 40.01249805,  11.18033989,  11.40175425],
           [ 42.3792402 ,  17.02938637,  16.2788206 ],
           [ 59.39696962,  43.01162634,  42.05948169],
           [ 55.97320788,  41.77319715,  40.79215611],
           [ 17.69180601,  20.80865205,  20.24845673],
           [ 41.72529209,  28.01785145,  27.01851217],
           [ 20.80865205,  44.01136217,  43.65775991],
           [ 65.9241989 ,  66.48308055,  65.520989  ]])

您可以通过此技巧计算最接近的质心的索引(为了便于阅读，它们分为3行)，

And you can compute the indices of the nearest centroids by this trick (they are split into 3 lines for readability),

    In: t0 = centroids[None, :, :] - dataPoints[:, None, :]
    In: t1 = np.linalg.norm(t0, axis=-1)
    In: t2 = np.argmin(t1, axis=-1)

现在t2具有索引，

    array([1, 2, 2, 2, 0, 2, 0, 2])

要找到#1群集，请使用布尔掩码t2 == 0

To find the #1 cluster, use the boolean mask t2 == 0,

    In: dataPoints[t2 == 0]
    Out: array([[16, 32],
                [20, 56]])

    In: dataPoints[t2 == 1]
    Out: array([[2, 4]])

    In: dataPoints[t2 == 2]
    Out: array([[17,  4],
                [45,  2],
                [45,  7],
                [32, 14],
                [68, 33]])

或者只是根据您的情况计算平均值，

Or just calculate the average in your case,

    In: np.mean(dataPoints[t2 == 0], axis=0)
    Out: array([ 18.,  44.])

    In: np.mean(dataPoints[t2 == 1], axis=0)
    Out: array([ 2.,  4.])

    In: np.mean(dataPoints[t2 == 2], axis=0)
    Out: array([ 41.4,  12. ])

当然，如果需要，可以在for循环中重写后面的块.

Of course, the latter blocks can be rewritten in for-loop if you want.

在我看来，按照numpy的惯例来制定解决方案可能是一个好习惯.

It might be a good practice to formulate the solution by numpy's conventions in my opinion.

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