如何指定numpy矩陣的子集來放置較小的矩陣？

Question

我有具有2的向下採樣率也就是說，我的金字塔的底部是一個形狀(256, 256)，其中一個新的水平是(128, 128)的圖像等圖像金字塔

我的目標是顯示該金字塔合併爲一個圖像。第一張圖片放在左邊。第二個放在右上角。每個後續圖像必須放置在先前的下方並楔入角落。

這裏是我當前的功能：

def pyramid2img(pmd): 
    ''' 
    Given a pre-constructed pyramid, this is a helper 
    function to display the pyramid in a single image. 
    ''' 

    # orignal shape (pyramid goes from biggest to smallest) 
    org_img_shp = pmd[0].shape 

    # the output will have to have 1.5 times the width 
    out_shp = tuple(int(x*y) \ 
     for (x,y) in zip(org_img_shp, (1, 1.5))) 
    new_img = np.zeros(out_shp, dtype=np.int8) 

    # i keep track of the top left corner of where I want to 
    # place the current image matrix 
    origin = [0, 0] 
    for lvl, img_mtx in enumerate(pmd): 

     # trying to specify the subset to place the next img_mtx in 
     sub = new_img[origin[0]:origin[0]+pmd[lvl].shape[0], 
      origin[1]:origin[1]+pmd[lvl].shape[1]]# = img_mtx 

     # some prints to see exactly whats being called above^
     print 'level {}, sub {}, mtx {}'.format(
      lvl, sub.shape, img_mtx.shape) 
     print 'sub = new_img[{}:{}, {}:{}]'.format(
      origin[0], origin[0]+pmd[lvl].shape[0], 
      origin[1], origin[1]+pmd[lvl].shape[1]) 

     # first shift moves the origin to the right 
     if lvl == 0: 
      origin[0] += pmd[lvl].shape[0] 
     # the rest move the origin downward 
     else: 
      origin[1] += pmd[lvl].shape[1] 

    return new_img 

輸出打印語句：

level 0, sub (256, 256), mtx (256, 256) 
sub = new_img[0:256, 0:256] 


level 1, sub (0, 128), mtx (128, 128) 
sub = new_img[256:384, 0:128] 


level 2, sub (0, 64), mtx (64, 64) 
sub = new_img[256:320, 128:192] 


level 3, sub (0, 32), mtx (32, 32) 
sub = new_img[256:288, 192:224] 


level 4, sub (0, 16), mtx (16, 16) 
sub = new_img[256:272, 224:240] 


level 5, sub (0, 8), mtx (8, 8) 
sub = new_img[256:264, 240:248] 


level 6, sub (0, 4), mtx (4, 4) 
sub = new_img[256:260, 248:252] 

如果你查看輸出，你可以看到，我想引用一個2D片的輸出圖像，以便我可以將金字塔的下一層放置在其中。

問題是，我正在執行的切片沒有給出我期望的形狀的2D數組。它認爲我試圖將（n，n）矩陣放入（0，n）矩陣中。

爲什麼當我指定像new_img[256:320, 128:192]片，它返回一個對象，具有形狀(0, 64)，不(64, 64)？

有沒有更簡單的方法來做我想做的事情？

Answer 1

這裏是一個例子。

首先創建一個金字塔：

import numpy as np 
import pylab as pl 
import cv2 

img = cv2.imread("earth.jpg")[:, :, ::-1] 

size = 512 
imgs = [] 
while size >= 2: 
    imgs.append(cv2.resize(img, (size, size))) 
    size //= 2 

這裏是合併圖像的代碼：

def align(size, width, loc): 
    if loc in ("left", "top"): 
     return 0 
    elif loc in ("right", "bottom"): 
     return size - width 
    elif loc == "center": 
     return (size - width) // 2 

def resize_canvas(img, shape, loc, fill=255): 
    new_img = np.full(shape + img.shape[2:], fill, dtype=img.dtype) 
    y = align(shape[0], img.shape[0], loc[0]) 
    x = align(shape[1], img.shape[1], loc[1]) 
    new_img[y:y+img.shape[0], x:x+img.shape[1], ...] = img 
    return new_img 

def vbox(imgs, align="right", fill=255): 
    width = max(img.shape[1] for img in imgs) 
    return np.concatenate([ 
      resize_canvas(img, (img.shape[0], width), ("top", align), fill=fill) 
      for img in imgs 
     ]) 

def hbox(imgs, align="top", fill=255): 
    height = max(img.shape[0] for img in imgs) 
    return np.concatenate([ 
      resize_canvas(img, (height, img.shape[1]), (align, "left"), fill=fill) 
      for img in imgs 
     ], axis=1) 

輸出：

pl.imshow(hbox([imgs[0], vbox(imgs[1:])]))