slice | 易学教程

How to operate on 2 mutable slices of a Rust array?

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问题 I have a function that needs to operate on two parts of a single array. The purpose is to be able to build an #[nostd] allocator that can return a variable slice of a bigger array to the caller and hang on to the remainder of the array for future allocations. Here's example code that fails: fn split<'a>(mut item: &'a mut [i32], place: usize) -> (&'a mut [i32], &'a mut [i32]) { (&mut item[0..place], &mut item[place..]) } fn main() { let mut mem: [i32; 2048] = [1; 2048]; let (mut array0, mut

Fast Way to slice image into overlapping patches and merge patches to image

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问题 Trying to slice a grayscale image of size 100x100 into patches of size 39x39 which are overlapping, with a stride-size of 1. That means that the next patch which starts one pixel to the right/or below is only different to the previous patch in one additional column/or row. Rough Outline of the code : First compute the indices for each patch, so to be able to construct the 2D array of patches from an image and to be able to construct an image from patches: patches = imgFlat[ind] 'patches' is a

Python plot - stacked image slices

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问题 I have a series of basic 2D images (3 for simplicity for now) and these are related to each other, analogous to frames from a movie: Within python how may I stack these slices on top of each other, as in image1->image2->image-3? I'm using pylab to display these images. Ideally an isometric view of the stacked frames would be good or a tool allowing me to rotate the view within code/in rendered image. Any assistance appreciated. Code and images shown: from PIL import Image import pylab

How can I slice each element of a numpy array of strings?

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问题 Numpy has some very useful string operations, which vectorize the usual Python string operations. Compared to these operation and to pandas.str , the numpy strings module seems to be missing a very important one: the ability to slice each string in the array. For example, a = numpy.array(['hello', 'how', 'are', 'you']) numpy.char.sliceStr(a, slice(1, 3)) >>> numpy.array(['el', 'ow', 're' 'ou']) Am I missing some obvious method in the module with this functionality? Otherwise, is there a fast

Slicing a list into n nearly-equal-length partitions [duplicate]

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问题 This question already has answers here : Splitting a list into N parts of approximately equal length (28 answers) Closed 2 years ago . I'm looking for a fast, clean, pythonic way to divide a list into exactly n nearly-equal partitions. partition([1,2,3,4,5],5)->[[1],[2],[3],[4],[5]] partition([1,2,3,4,5],2)->[[1,2],[3,4,5]] (or [[1,2,3],[4,5]]) partition([1,2,3,4,5],3)->[[1,2],[3,4],[5]] (there are other ways to slice this one too) There are several answers in here Iteration over list slices

ICE使用总结

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客户端的slice-to-c++映射 slice标识符会映射到相同的c++标识符，如果一个slice的标识符与某个c++关键字一样，那么对应的c++ 标识符会加上前缀 _cpp_ 命名空间的映射是一样的接口映射：要调用一个远地操作，要调用本地类实例的成员函数，这个实例代表的是远地的对象。这使得映射变得容易，使用也直观。举例： interface Simple { void op(); }; 例如上述slice接口，映射到c++ namespace IceProxy { class Simple : public virtual IceProxy::Ice::Object { public: void op(); void op(const Ice::Context &); // ... }; } typedef IceInternal::ProxyHandle<IceProxy::Simple> SimplePrx; 如上所示，上述slice生成一个类simple和一个类代理SimplePrx c++程序会一般直接使用这个代理句柄，而不会使用这个类操作的映射对于接口的每一个操作，代理类都有一个对应的同名成员函数，要调用某一个操作，要通过代理句柄调用它。例如，对于接口： module Filesystem { interface Node { nonmutating

托马斯微积分从入门到失望

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决定把例题用程序都完成一遍。从最基本的开始:语言选择用python,vex,Houdini作图 <1> a,求球的体积.半径为4,中心点为0，左断点为-4,右断点为4 import math radius = 4.000 diameter = radius *2 # sphere r=4 R=8 # this spere is y=sqrt(16-x*x) # per cylinder volume is PI*r*r*dtx def sphere_function(xpos,dtx): return math.sqrt(16.00-xpos*xpos) * math.sqrt(16.00-xpos*xpos)*dtx # @n is the is this sphere that will split to n piece along xpos # if sphere r=4, slice is 8, dtx= R/8 = 1 def calculateVolume(n): v = 0 dtx = diameter/n for j in range(0,n,1): xp = j*(diameter/n) - radius #if sphere radius is 4,left plot is -4 ,right plot is 4 v += sphere_function(xp

go slice

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package main import ( "fmt" ) type Pair struct { a int b int } type PairAndFreq struct { Pair Freq int } type PairSlice []PairAndFreq type PairSliceSlice []PairSlice func (pss PairSliceSlice) Weed() { fmt.Println(pss[0]) weed(pss[0]) fmt.Println(pss[0]) } func weed(ps PairSlice) { m := make(map[Pair]int) for _, v := range ps { m[v.Pair]++ } ps = ps[:0] for k, v := range m { ps = append(ps, PairAndFreq{k, v}) } fmt.Println(ps) } func main() { pss := make(PairSliceSlice, 12) pss[0] = PairSlice{PairAndFreq{Pair{1, 1}, 1}, PairAndFreq{Pair{1, 1}, 1}} pss.Weed() } 这段代码最后输出的是 {1 1} 1} {{1 1} 1}] [{

pytorch中tensorboardX进行可视化

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环境依赖： pytorch 0.4以上 tensorboardX: pip install tensorboardX、pip install tensorflow 在项目代码中加入tensorboardX的记录代码，生成文件并返回到浏览器中显示可视化结果。官方示例：默认设置是在根目录下生成一个runs文件夹，里面存储summary的信息。在runs的同级目录下命令行中输入： tensorboard --logdir runs (不是输tensorboardX) 会出来一个网站,复制到浏览器即可可视化loss,acc,lr等数据的变化过程. 举例说明pytorch中设置summary的方式: 1 import argparse 2 import os 3 import numpy as np 4 from tqdm import tqdm 5 6 from mypath import Path 7 from dataloaders import make_data_loader 8 from modeling.sync_batchnorm.replicate import patch_replication_callback 9 from modeling.deeplab import * 10 from modeling.psp_net import * 11 from

Is there a way to get slice as result of Find()?

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问题 Now I'm doing: sess := mongodb.DB("mybase").C("mycollection") var users []struct { Username string `bson:"username"` } err = sess.Find(nil).Select(bson.M{"username": 1, "_id": 0}).All(&users) if err != nil { fmt.Println(err) } var myUsers []string for _, user := range users{ myUsers = append(myUsers, user.Username) } Is there a more effective way to get slice with usernames from Find (or another search function) directly, without struct and range loop? 回答1: The result of a MongoDB find() is