如何將多個(gè)圖像填充到包含它們的最小形狀？

Python

守著星空守著你 2022-06-07 17:59:04

我正在嘗試編寫一個(gè)函數(shù)來(lái)加載和處理 NN 的數(shù)據(jù)。作為輸入，我有一組不同尺寸的圖片。圖片應(yīng)表示為具有 RGB 通道的 3D numpy 數(shù)組。我需要它們具有相同的大小（最大圖片的大?。Ｎ乙呀?jīng)嘗試過(guò)np.pad，但似乎我不知道它應(yīng)該如何工作。實(shí)際上，即使我得到了填充，我也不知道如何根據(jù)圖片的大小來(lái)改變它。這是代碼：from PIL import Imageimport numpy as npimport cv2import osdef load_data(path): aminoacids = ['Ala','Arg','Asn','Asp','Cys','Gln','Glu','Gly','His','Ile', 'Ini', 'Leu','Lys','Met','Phe','Pro','Pyr', 'Sec','Ser','Thr','Trp','Tyr','Val'] matrix = [] answer_labeled = [] names = os.listdir(path) for i in names: matrix = cv2.imread(path + i, 1) matrix = np.pad(matrix, (0, 1), 'constant', constant_values=[255,255,255]) for y in aminoacids: if y + "-" in i: a = [matrix, y] answer_labeled.append(a) return answer_labeleddata_processed = load_data("/content/drive/My Drive/Thesis/dl/img/ans/")我收到此錯(cuò)誤：ValueErrorTraceback (most recent call last)<ipython-input-50-e021738e59ea> in <module>() 20 return answer_labeled 21 ---> 22 data_processed = load_data("/content/drive/My Drive/Thesis/dl/img/ans/") 23 24 # print(len(os.listdir("/content/drive/My Drive/Thesis/dl/img/ans/")))<ipython-input-50-e021738e59ea> in load_data(path) 13 for i in names: 14 matrix = cv2.imread(path + i, 1)---> 15 matrix = np.pad(matrix, (0, 1), 'constant', constant_values=[255,255,255]) 16 for y in aminoacids: 17 if y + "-" in i:/usr/local/lib/python2.7/dist-packages/numpy/lib/arraypad.pyc in pad(array, pad_width, mode, **kwargs) 1208 kwargs[i] = _as_pairs(kwargs[i], narray.ndim, as_index=True) 1209 if i in ['end_values', 'constant_values']:-> 1210 kwargs[i] = _as_pairs(kwargs[i], narray.ndim) 1211 else: 1212 # Drop back to old, slower np.apply_along_axis mode for user-supplied當(dāng)然，我試圖用谷歌搜索這個(gè)錯(cuò)誤，但沒有找到對(duì)我有用或可以理解的東西（因?yàn)槲艺娴氖蔷幊绦率郑?。我將不勝感激任何幫助和想法?

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2 回答

繁星coding

TA貢獻(xiàn)1797條經(jīng)驗(yàn) 獲得超4個(gè)贊

我曾經(jīng)不得不解決一個(gè)類似的任務(wù)，所以我為它創(chuàng)建了以下函數(shù)。它允許為每個(gè)維度指定大小差異的分?jǐn)?shù)，應(yīng)該在之前和之后填充（類似于np.pad）。例如，如果您有兩個(gè)形狀數(shù)組(3,)和(5,)，那么before=1將在左側(cè)填充整個(gè)差異（在本例2中為），而在左側(cè)填充before=0.5一個(gè)元素，在右側(cè)填充一個(gè)元素。與np.pad這些因素類似，也可以為每個(gè)維度指定。這是實(shí)現(xiàn)：

import numpy as np

def pad_max_shape(arrays, before=None, after=1, value=0, tie_break=np.floor):

"""Pad the given arrays with a constant values such that their new shapes fit the biggest array.

Parameters

----------

arrays : sequence of arrays of the same rank

before, after : {float, sequence, array_like}

Similar to `np.pad -> pad_width` but specifies the fraction of values to be padded before

and after respectively for each of the arrays. Must be between 0 and 1.

If `before` is given then `after` is ignored.

value : scalar

The pad value.

tie_break : ufunc

The actual number of items to be padded _before_ is computed as the total number of elements

to be padded times the `before` fraction and the actual number of items to be padded _after_

is the remainder. This function determines how the fractional part of the `before` pad width

is treated. The actual `before` pad with is computed as ``tie_break(N * before).astype(int)``

where ``N`` is the total pad width. By default `tie_break` just takes the `np.floor` (i.e.

attributing the fraction part to the `after` pad width). The after pad width is computed as

``total_pad_width - before_pad_width``.

Returns

-------

padded_arrays : list of arrays

Notes

-----

By default the `before` pad width is computed as the floor of the `before` fraction times the number

of missing items for each axis. This is done regardless of whether `before` or `after` is provided

as a function input. For that reason the fractional part of the `before` pad width is attributed

to the `after` pad width (e.g. if the total pad width is 3 and the left fraction is 0.5 then the

`before` pad width is 1 and the `after` pad width is 2; in order to f). This behavior can be controlled

with the `tie_break` parameter.

"""

shapes = np.array([x.shape for x in arrays])

if before is not None:

before = np.zeros_like(shapes) + before

else:

before = np.ones_like(shapes) - after

max_size = shapes.max(axis=0, keepdims=True)

margin = (max_size - shapes)

pad_before = tie_break(margin * before.astype(float)).astype(int)

pad_after = margin - pad_before

pad = np.stack([pad_before, pad_after], axis=2)

return [np.pad(x, w, mode='constant', constant_values=value) for x, w in zip(arrays, pad)]

對(duì)于您的示例，您可以按如下方式使用它：

test = [np.ones(shape=(i, i, 3)) for i in range(5, 10)]

result = pad_max_shape(test, before=0.5, value=255)

print([x.shape for x in result])

print(result[0][:, :, 0])

這會(huì)產(chǎn)生以下輸出：

[(9, 9, 3), (9, 9, 3), (9, 9, 3), (9, 9, 3), (9, 9, 3)]

[[255. 255. 255. 255. 255. 255. 255. 255. 255.]

[255. 255. 255. 255. 255. 255. 255. 255. 255.]

[255. 255. 1. 1. 1. 1. 1. 255. 255.]

[255. 255. 255. 255. 255. 255. 255. 255. 255.]

[255. 255. 255. 255. 255. 255. 255. 255. 255.]]

所以我們可以看到每個(gè)數(shù)組都被對(duì)稱地填充到最大數(shù)組的形狀(9, 9, 3)。

反對(duì) 回復(fù) 2022-06-07

米脂

TA貢獻(xiàn)1836條經(jīng)驗(yàn) 獲得超3個(gè)贊

的使用np.pad()實(shí)際上是有據(jù)可查的。

一個(gè)適用于您提供的數(shù)字的 3D 數(shù)據(jù)的示例是：

import numpy as np

arr = np.random.randint(0, 255, (72, 72, 3))

new_arr = np.pad(

arr, ((0, 92 - 72), (0, 92 - 72), (0, 0)),

'constant', constant_values=255)

print(new_arr.shape)

# (92, 92, 3)

編輯

要解決完整的問(wèn)題，您需要首先確定最大尺寸，然后相應(yīng)地填充所有其他圖像。FlyingCircus包為您提供了許多功能，可以更輕松地完成工作（免責(zé)聲明：我是它的主要作者）。

如果您可以將所有圖像放入內(nèi)存中，最簡(jiǎn)單的方法就是使用fc.extra.multi_reframe()，即：

import flyingcircus as fc

new_arrs = fc.extra.multi_reframe(arrs, background=255)

如果您無(wú)法將所有數(shù)據(jù)放入內(nèi)存中，則應(yīng)分兩次執(zhí)行此操作，一次計(jì)算適合所有輸入的最小形狀，然后執(zhí)行實(shí)際填充fc.extra.reframe()：

# assume your data is loaded with `load(filepath)`

# ... and saved with `save(array, filepath)`

# : first pass

shapes = [load(filepath).shape for filepath in filepaths]

target_shape = tuple(np.max(np.array(shapes), axis=0))

# : second pass

for filepath in filepaths:

arr = load(filepath)

new_arr = fc.extra.reframe(arr, target_shape, 0.5, 255)

save(new_arr, filepath)

在內(nèi)部，fc.extra.reframe()正在使用np.pad()（或類似但更快的東西），它大致相當(dāng)于：

def reframe(arr, target_shape, position=0.5, *args, **kws):

source_shape = arr.shape

padding = tuple(

(int(position * (dim_target - dim_source)),

(dim_target - dim_source) - int(position * (dim_target - dim_source)))

for dim_target, dim_source in zip(target_shape, source_shape))

return np.pad(arr, padding, *args, **kws)

reframe(arr, target_shape, 0.5, 'constant', constant_values=255)

請(qǐng)注意，position參數(shù)決定了數(shù)組相對(duì)于新形狀的位置。的默認(rèn)值0.5會(huì)將所有圖像放在中心，而0.0或1.0會(huì)將其推到所有軸上新形狀的一側(cè)或另一側(cè)。它的 FlyingCircus 版本更加靈活，您可以position分別為所有軸指定一個(gè)值。

反對(duì) 回復(fù) 2022-06-07