slice | 易学教程

NumPy slice notation in a dictionary

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问题 I wonder if it is possible to store numpy slice notation in a python dictionary. Something like: lookup = {0:[:540], 30:[540:1080], 60:[1080:]} It is possible to use native python slice syntax, e.g. slice(0,10,2) , but I have not been able to store more complex slices. For example, something that is multidimensional [:,:2,:, :540] . My current work around is to store the values as tuples and then unpack these into the necessary slices. Working in Python 2.x. 回答1: The syntax [:, :2, :, :540]

Array.prototype.slice.call()

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MDN中对于 Array.prototype.slice.() 的介绍中，提到了类数组对象。以下是原文： slice 方法可以用来将一个类数组（Array-like）对象/集合转换成一个新数组。你只需将该方法绑定到这个对象上。一个函数中的 arguments 就是一个类数组对象的例子。 function list() { return Array.prototype.slice.call(arguments); } var list1 = list(1, 2, 3); // [1, 2, 3] 除了使用 Array.prototype.slice.call(arguments) ，你也可以简单的使用 [].slice.call(arguments) 来代替。所以arguments并不是真正的数组对象，只是与数组类似而已，所以它并没有slice这个方法，而Array.prototype.slice.call(arguments, 1)可以理解成是让arguments转换成一个数组对象，让arguments具有slice()方法。同理可知，我们可以给 Array.prototype.slice.call(arguments) 加上第二个参数。 function list() { return Array.prototype.slice.call(arguments, 1); }

Could anyone explain this strange behaviour of appending to golang slices

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问题 The program below has unexpected output. func main(){ s:=[]int{5} s=append(s,7) s=append(s,9) x:=append(s,11) y:=append(s,12) fmt.Println(s,x,y) } output: [5 7 9] [5 7 9 12] [5 7 9 12] Why is the last element of x 12 ? 回答1: A slice is only a window over part of an array, it has no specific storage. This means that if you have two slices over the same part of an array, both slices must "contain" the same values. Here's exactly what happens here : When you do the first append , you get a new

Python : Why use “list[:]” when “list” refers to same thing?

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问题 Consider a list >>> l=[1,2,3] . What is the benefit of using >>> l[:] when >>> l prints the same thing as former does? Thanks. 回答1: It creates a (shallow) copy. >>> l = [1,2,3] >>> m = l[:] >>> n = l >>> l.append(4) >>> m [1, 2, 3] >>> n [1, 2, 3, 4] >>> n is l True >>> m is l False 回答2: l[:] is called slice notation. It can be used to extract only some of the elements in the list, but in this case the bounds are omitted so the entire list is returned, but because of the slice, this will

What is the purpose of the two colons in this Python string-slicing statement?

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问题 For example, str = "hello" str[1::3] And where can I find this in Python documentation? 回答1: in sequences' description: s[i:j:k] slice of s from i to j with step k The slice of s from i to j with step k is defined as the sequence of items with index x = i + n*k such that 0 <= n < (j-i)/k . In other words, the indices are i , i+k , i+2*k , i+3*k and so on, stopping when j is reached (but never including j ). If i or j is greater than len(s) , use len(s ). If i or j are omitted or None , they

Strange behavior with python slicing [duplicate]

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问题 This question already has answers here : Reversing a list slice in python (2 answers) Closed 19 days ago . Suppose we have this list: >>> a = [x for x in range(10)] >>> print(a) [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] Separately, both ways to slice work as expected: >>> a[3:8] [3, 4, 5, 6, 7] >>> a[::-1] [9, 8, 7, 6, 5, 4, 3, 2, 1, 0] But, when combined: >>> a[3:8:-1] [] I would expect it to be [7, 6, 5 ,4, 3] or perhaps [6, 5, 4, 3, 2] (if reversing happened first). It is also interesting to consider

Ruby String: how to match a Regexp from a defined position

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问题 I want to match a regexp from a ruby string only from a defined position. Matches before that position do not interest me. Moreover, I'd like \A to match this position. I found this solution: code[index..-1][/\A[a-z_][a-zA-Z0-9_]*/] This match the regexp at position index in the string code . If the match is not exactly at position index , it return nil . Is there a more elegant way to do this (I want to avoid to create the temporary string with the first slice)? Thanks 回答1: You could use ^.{

Creating a function that takes an equation given as a string and computes it [duplicate]

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问题 This question already has answers here : Evaluating a mathematical expression in a string (11 answers) Closed 4 years ago . As part of an assignment, I am creating a function that takes in a string, which is an equation. here is an example of one: 48+6x6/3=6x8-9x2. The compute function takes one side of the equal sign and evaluates it. I am not too concerned with splitting the equation. I believe I can just slice it with s[:s.find("=")] . My main problem is the compute function itself. I will

Creating a function that takes an equation given as a string and computes it [duplicate]

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why can't I use reflection to take the address of a slice?

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问题 How come this works: slice := make([]string, 0, 10) sliceptr := &slice this too: sliceptr := &[]string{"foo","bar","baz"} But this doesn't: sliceaddrval := reflect.ValueOf([]string{"foo","bar","baz"}).Addr() It panics with: reflect.Value.Addr of unaddressable value EDIT: Overall what I'm trying to do is take a struct that is of an unknown type, make a slice of structs of that type and return a pointer to it (I'm using github.com/jmoiron/modl which requires a pointer to slice to populate with