問題描述

itertools python 模塊為迭代器實(shí)現(xiàn)了一些基本構(gòu)建塊.正如他們所說，它們形成了一個迭代器代數(shù)".我期待，但我找不到使用該模塊進(jìn)行以下迭代的簡潔方法.給定一個有序?qū)崝?shù)列表，例如

The itertools python module implements some basic building blocks for iterators. As they say, "they form an iterator algebra". I was expecting, but I could not find a succinctly way of doing the following iteration using the module. Given a list of ordered real numbers, for example

a = [1.0,1.5,2.0,2.5,3.0]

...返回一個新列表(或只是迭代)按一些 n 值分組，比如 2

... return a new list (or just iterate) grouping by some n value, say 2

b = [(1.0,1.5),(1.5,2.0),(2.0,2.5),(2.5,3.0)]

我發(fā)現(xiàn)這樣做的方法如下.首先將列表一分為二，帶有偶數(shù)和賠率索引:

The way I found of doing this was as follows. First split the list in two, with evens and odds indexes:

even, odds = a[::2], a[1::2]

然后構(gòu)造新列表:

b = [(even, odd) for even, odd in zip(evens, odds)]
b = sorted(b + [(odd, even) for even, odd in zip(evens[1:], odds)])

本質(zhì)上，它類似于移動均值.

In essence, it is similar to a moving mean.

是否有簡潔的方法(使用或不使用 itertools)?

PS:

應(yīng)用程序

將 a 列表想象為實(shí)驗(yàn)期間發(fā)生的某些事件的時間戳集:

Imagine the a list as the set of timestamps of some events occurred during an experiment:

timestamp       event
47.8            1a
60.5            1b
67.4            2a
74.5            2b
78.5            1a
82.2            1b
89.5            2a
95.3            2b
101.7           1a
110.2           1b
121.9           2a
127.1           2b

...

此代碼用于根據(jù)不同的時間窗口對這些事件進(jìn)行分段.現(xiàn)在我對 2 連續(xù)事件之間的數(shù)據(jù)感興趣；'n > 2' 僅用于探索目的.

This code is being used to segment those events in accord with different temporal windows. Right now I am interested in the data between 2 successive events; 'n > 2' would be used only for exploratory purposes.

推薦答案

對于2，你可以這樣做

b = zip(a, a[1:])  # or list(zip(...)) on Python 3 if you really want a list

對于固定的 n，技術(shù)類似:

For fixed n, the technique is similar:

# n = 4
b = zip(a, a[1:], a[2:], a[3:])

對于變量 n，您可以壓縮可變數(shù)量的切片，或者(尤其是當(dāng)窗口大小接近 a 的大小時)您可以使用切片直接獲取窗口:

For variable n, you could zip a variable number of slices, or (especially if the window size is close to the size of a) you could use slicing to take windows directly:

b = zip(*[a[i:] for i in xrange(n)])
# or
b = [tuple(a[i:i+n]) for i in xrange(len(a)-n+1)]

如果 a 不是列表，您可以從 itertools 文檔中概括 pairwise 配方:

If a is not a list, you could generalize the pairwise recipe from the itertools docs:

import copy
import itertools

def nwise(iterable, n):
    # Make n tees at successive positions along the iterable.
    tees = list(itertools.tee(iterable, 1))
    for _ in xrange(n-1):
        tees.append(copy.copy(tees[-1]))
        next(tees[-1])

    return zip(*tees)

這篇關(guān)于迭代列表的 n 個連續(xù)元素(重疊)的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網(wǎng)！