flatten

问题 In the context of a discussion in the comments of this question it was mentioned that while concatenating a sequence of strings simply takes ''.join([str1, str2, ...]) , concatenating a sequence of lists would be something like list(itertools.chain(lst1, lst2, ...)) , although you can also use a list comprehension like [x for y in [lst1, lst2, ...] for x in y] . What surprised me is that the first method is consistently faster than the second: import random import itertools random.seed(100)

问题 Introduction R code is written by using Sparklyr package to create database schema. [Reproducible code and database is given] Existing Result root |-- contributors : string |-- created_at : string |-- entities (struct) | |-- hashtags (array) : [string] | |-- media (array) | | |-- additional_media_info (struct) | | | |-- description : string | | | |-- embeddable : boolean | | | |-- monetizable : bollean | | |-- diplay_url : string | | |-- id : long | | |-- id_str : string | |-- urls (array) |-

问题 I have a nested tuple structure like (String,(String,Double)) and I want to transform it to (String,String,Double) . I have various kinds of nested tuple, and I don't want to transform each manually. Is there any convenient way to do that? 回答1: If you use shapeless, this is exactly what you need, I think. 回答2: There is no flatten on a Tupple. But if you know the structure, you can do something like this: implicit def flatten1[A, B, C](t: ((A, B), C)): (A, B, C) = (t._1._1, t._1._2, t._2)

问题 I have been flattening domain objects into DTOs as shown in the example below: public class Root { public string AParentProperty { get; set; } public Nested TheNestedClass { get; set; } } public class Nested { public string ANestedProperty { get; set; } } public class Flattened { public string AParentProperty { get; set; } public string ANestedProperty { get; set; } } // I put the equivalent of the following in a profile, configured at application start // as suggested by others: Mapper

问题 Is there any easy way to flatten import numpy np.arange(12).reshape(3,4) Out[]: array([[ 0, 1, 2, 3], [ 4, 5, 6, 7], [ 8, 9, 10, 11]]) into array([ 0, 1, 4, 5, 8, 9, 2, 3, 6, 7, 10, 11]) 回答1: It seems like you are looking to consider a specific number of cols to form blocks and then getting the elements in each block and then moving onto the next ones. So, with that in mind, here's one way - In [148]: a Out[148]: array([[ 0, 1, 2, 3], [ 4, 5, 6, 7], [ 8, 9, 10, 11]]) In [149]: ncols = 2 # no.

问题 Can someone help me to break down exactly the order of execution for the following versions of flatten? I'm using Racket. version 1, is from racket itself, while version two is a more common? implementation. (define (flatten1 list) (let loop ([l list] [acc null]) (printf "l = ~a acc = ~a\n" l acc) (cond [(null? l) acc] [(pair? l) (loop (car l) (loop (cdr l) acc))] [else (cons l acc)]))) (define (flatten2 l) (printf "l = ~a\n" l) (cond [(null? l) null] [(atom? l) (list l)] [else (append

问题 I want to push all individual elements of a source array onto a target array, target.push(source); puts just source's reference on the target list. In stead I want to do: for (i = 0; i < source.length; i++) { target.push(source[i]); } Is there a way in javascript to do this more elegant, without explicitly coding a repetition loop? And while I'm at it, what is the correct term? I don't think that "flat push" is correct. Googling did not yield any results as source and target are both arrays.

问题 I am surprised that no one on StackOverflow asked this question before. Looking through the JSON object documentation and a quick google search did not yield satisfactory results. What's the advantage of it? How does it work? Edit: To make it clear, take a look at this flatten/un-flatten example. Fastest way to flatten / un-flatten nested JSON objects Thank you. 回答1: There are many situations where you get JSON text that was automatically built by some library. Throughout the programming

问题 I've already searched SO for how to flatten a list of lists (i.e. here:Making a flat list out of list of lists in Python) but none of the solutions I find addresses flattening a list of lists of lists to just a list of lists. I have: my_list = [ [ [1,2,3],[4,5] ], [ [9],[8,9,10],[3,4,6] ], [ [1] ] ] I want: my_list = [ [1,2,3,4,5], [9,8,9,10,3,4,6], [1] ] The solution should work for a list of floats as well. Any suggestions? 回答1: If we apply the logic from this answer, should not it be just:

问题 Given the following list of tuples: INPUT = [(1,2),(1,),(1,2,3)] How would I flatten it into a list? OUTPUT ==> [1,2,1,1,2,3] Is there a one-liner to do the above? Similar: Flatten list of Tuples in Python 回答1: You could use a list comprehension: >>> INPUT = [(1,2),(1,),(1,2,3)] >>> [y for x in INPUT for y in x] [1, 2, 1, 1, 2, 3] >>> itertools.chain.from_iterable is also used a lot in cases like this: >>> from itertools import chain >>> INPUT = [(1,2),(1,),(1,2,3)] >>> list(chain.from