![](/img/trans.png)
[英]how to get subset of matrix in class numpy.matrix.defmatrix.matrix
[英]How can I specify a subset of a numpy matrix for placement of smaller matrix?
我有一個向下采樣率為2的圖像金字塔。也就是說,我的金字塔的底部是形狀為(256, 256)
的圖像,下一個級別是(128, 128)
等。
我的目標是將此金字塔顯示為單個圖像。 第一張圖片位於左側。 第二個放置在右上角。 每個后續圖像必須放置在前一個圖像的下方並楔入角落。
這是我當前的功能:
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
打印聲明的輸出:
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]
如果查看輸出,可以看到我正在嘗試引用輸出圖像的2切片,以便將金字塔的下一層放置在其中。
問題是我執行的切片沒有給出我期望的形狀的2d數組。 它認為我正在嘗試將(n,n)矩陣放入(0,n)矩陣中。
當我指定像new_img[256:320, 128:192]
這樣的切片時,為什么返回形狀為(0, 64)
new_img[256:320, 128:192]
(0, 64)
,NOT (64, 64)
呢?
有沒有更簡單的方法可以做我想做的事情?
這是一個例子。
首先創建金字塔:
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
這是合並圖像的代碼:
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:])]))
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.