Swift相当于Python切片赋值

Question

在Python中,可以有一个列表(类似于swift中的数组):

>>> li=[0,1,2,3,4,5]

并在列表的任何/所有列表上执行切片分配:

>>> li[2:]=[99]         # note then end index is not needed if you mean 'to the end'
>>> li
[0, 1, 99]

Swift有一个类似的切片赋值(这在swift交互式shell中):

  1> var arr=[0,1,2,3,4,5]
arr: [Int] = 6 values {
  [0] = 0
  [1] = 1
  [2] = 2
  [3] = 3
  [4] = 4
  [5] = 5
}
  2> arr[2...arr.endIndex-1]=[99]
  3> arr
$R0: [Int] = 3 values {
  [0] = 0
  [1] = 1
  [2] = 99
}

到现在为止还挺好.但是,有几个问题.

首先,swift不适用于空列表,或者索引是否在endIndex.如果切片索引在结束索引之后,则Python附加:

>>> li=[]             # empty
>>> li[2:]=[6,7,8]
>>> li
[6, 7, 8]
>>> li=[0,1,2]
>>> li[999:]=[999]
>>> li
[0, 1, 2, 999]

swift中的等价物是一个错误:

  4> var arr=[Int]()
arr: [Int] = 0 values
  5> arr[2...arr.endIndex-1]=[99]
fatal error: Can't form Range with end < start

这很容易测试和编码.

第二个问题是杀手:它的速度非常慢.考虑这个Python代码来执行浮点列表的精确求和:

def msum(iterable):
    "Full precision summation using multiple floats for intermediate values"
    # Rounded x+y stored in hi with the round-off stored in lo.  Together
    # hi+lo are exactly equal to x+y.  The inner loop applies hi/lo summation
    # to each partial so that the list of partial sums remains exact.
    # Depends on IEEE-754 arithmetic guarantees.  See proof of correctness at:
    # www-2.cs.cmu.edu/afs/cs/project/quake/public/papers/robust-arithmetic.ps

    partials = []               # sorted, non-overlapping partial sums
    for x in iterable:
        i = 0
        for y in partials:
            if abs(x) < abs(y):
                x, y = y, x
            hi = x + y
            lo = y - (hi - x)
            if lo:
                partials[i] = lo
                i += 1
            x = hi
        partials[i:] = [x]
    return sum(partials, 0.0)

它的工作原理是保持一个hi/lo部分求和,以便精确地msum([.1]*10)产生1.0而不是0.9999999999999999.C等价物msum是Python中数学库的一部分.

我试图在swift中复制:

func msum(it:[Double])->Double {
    // Full precision summation using multiple floats for intermediate values 
    var partials=[Double]()
    for var x in it {
        var i=0
        for var y in partials{
            if abs(x) < abs(y){
                (x, y)=(y, x)
            }
            let hi=x+y
            let lo=y-(hi-x)
            if abs(lo)>0.0 {
                partials[i]=lo
                i+=1
            }
            x=hi
        }
        // slow part trying to replicate Python's slice assignment partials[i:]=[x]
        if partials.endIndex>i {
            partials[i...partials.endIndex-1]=[x]
        }
        else {
            partials.append(x)
        }    
    } 
    return partials.reduce(0.0, combine: +)
}

测试功能和速度:

import Foundation
var arr=[Double]()
for _ in 1...1000000 {
    arr+=[10, 1e100, 10, -1e100]
    }

print(arr.reduce(0, combine: +))    // will be 0.0
var startTime: CFAbsoluteTime!
startTime = CFAbsoluteTimeGetCurrent()
print(msum(arr), arr.count*5)          // should be arr.count * 5
print(CFAbsoluteTimeGetCurrent() - startTime)

在我的机器上,需要7秒才能完成.Python原生msum需要2.2秒(大约快4倍),库fsum函数需要0.09秒(几乎快90倍)

我试图替换partials[i...partials.endIndex-1]=[x],arr.removeRange(i..<arr.endIndex)然后追加.快一点但不多.

题:

1. 这是惯用的快速: partials[i...partials.endIndex-1]=[x]
2. 有更快/更好的方式吗？

Answer 1

首先(正如在评论中已经说过的),Swift中的非优化代码和优化代码之间存在巨大差异("-Onone"vs"-O"编译器选项,或Debug与Release配置),因此对于性能测试make确保选中"Release"配置.(如果使用Instruments分析代码,"Release"也是默认配置).

使用半开范围有一些优点:

var arr = [0,1,2,3,4,5]
arr[2 ..< arr.endIndex] = [99]
print(arr) // [0, 1, 99]

事实上,这是一个怎样的范围在内部存储,它可以让你插入片末的数组(但不超过作为在Python):

var arr = [Int]()
arr[0 ..< arr.endIndex] = [99]
print(arr) // [99]

所以

if partials.endIndex > i {
    partials[i...partials.endIndex-1]=[x]
}
else {
    partials.append(x)
} 

相当于

 partials[i ..< partials.endIndex] = [x]
 // Or: partials.replaceRange(i ..< partials.endIndex, with: [x])

但是,这不是性能提升.似乎在Swift中替换切片很慢.截断数组并附加新元素

partials.replaceRange(i ..< partials.endIndex, with: [])
partials.append(x)

在我的计算机上将测试代码的时间从大约1.25秒减少到0.75秒.