nested-lists

问题 At first I have to say I'm new to AngularJS but I have to modify a web app, that is to nest list inside another: <ul class="first-level"> <li ng-repeat="item in list">{{item.parentNo1}} <ul class="level-2"> <li ng-repat="item in list">{{item.childOfCurrentParent}}</li> . . . </ul> </li> <li ng-repeat="item in list">{{item.parentNo2}} <ul class="level-2"> <li ng-repat="item in list">{{item.childOfCurrentParent}}</li> . . . </ul> </li> </ul> I received an array of objects (JSON) like this: [{

问题 I am a newbie in python, I have two lists: l1 = ['a','b','c','d'] l2 = ['new'] i want to get new list like this l3 = [('a','new'),('b','new'),('c','new'),('d','new')] What is the best way to combine the two lists? 回答1: >>> from itertools import product >>> l1 = ['a','b','c','d'] >>> l2 = ['new'] >>> list(product(l1,l2)) [('a', 'new'), ('b', 'new'), ('c', 'new'), ('d', 'new')] 回答2: If l2 always just has the one element there is no need to overcomplicate things l3 = [(x, l2[0]) for x in l1] 回答3

问题 Say I have a list x = [1, 2, 3, 4] Is there a recursive method where i can go through the list to find the value? I want to ultimately be able to compare a returned value in the list, (or nested list) to an arbitrary number to see it it matches. I can think a way to do this using a for loop, but i have trouble imagining a recursive method to do the same thing. I know that I can't set a counter to keep track of my position in the list because calling the function recursively would just reset

问题 class EntityCompositeId { private Long firstId; private Long secondId; // getter & setter... } class EntityComposite { private EntityCompositeId id; private String first; private String second; // getter & setter... } List<EntityComposite> listEntityComposite = .... Supose this content 1, 1, "firstA", "secondBirdOne" 1, 2, "firstA", "secondBirdTwo" 1, 3, "firstA", "secondBirdThree" 2, 1, "firstB", "secondCatOne" 2, 2, "firstB", "secondCatTwo" 2, 3, "firstB", "secondCatThree" 3, 1, "firstC",

问题 I have a list like that [[1]] [1] a1 b1 c1 [[2]] [1] a2 b2 c2 [[3]] [1] a3 b3 c3 I want specific element removed from each part of it: [[1]] [1] a1 c1 [[2]] [1] a2 c2 [[3]] [1] a3 c3 I tried tail but removes "outer" elements. Maybe some indexing would do? 回答1: Assuming the pattern is just that you want the second element removed, lapply(List, function(x) x[-2]) 回答2: Using purrr::map it would be even shorter by doing # setup some example data nestedList = list(list(4,5,6),list(1,2,3)) # remove

问题 I have this large corpus data in dataframe res (dataframe) text.1 1 <NA> 2 beren stuart vanuatu monday october venkatesh ramesh sandeep talanki nagaraj subject approve qlikview gpa access process form gpa access email requestor line manager access granted raj add user qlikview workgroup gpa access form requestors lim tek kon vanuatu address lini high port vila efate title relationship manager emerging corporates employee id lan id limtk bsbcc authorising manager beren stuart vanuatu read gpa

问题 How can I create a nested list of categories in Laravel? I want to create something like this: --- Php ------ Laravel --------- Version ------------ V 5.7 --- Python ------ Django --- Ruby .......... My table's fields are: id | name | parent_id If I have to add another column like depth, please tell me. I am using this following code, but I think it is not the best solution for creating nested-list of categories besides, I can not pass this function to my view. function rec($id) { $model =

问题 Problem: I have many nodes that need package updates. Some of the nodes have these packages installed and some do not. The goal is to 1. check if a package is installed using the yum module. 2. if package is installed and update is available then run yum update I know this is easily done via command or shell but very inefficient. tasks: - name: check if packages are installed yum: list="{{ item }}" with_items: - acpid - c-ares - automake register: packages - debug: var: packages will produce

问题 I have a complex list and need to select a subset from it, based on the value of a boolean element (I need records with hidden value equal to FALSE ). I've tried the following code, based on index vectors , but it fails (as shown at the end of this output): startups <- data$startups[data$startups$hidden == FALSE] Or, alternatively: startups <- data$startups[!as.logical(data$startups$hidden)] Interactive R session proves that the data is there: Browse[1]> str(data$startups, list.len=3) List of

问题 I have a flat list, for example: flat = ['1', '1-1', '1-1-1', '1-2', '2', '2-1', '2-2', '3'] that I need to convert to a nested list, where each level (dash followed by a number) starts a new sublist, for example: result = ['1', ['1-1', ['1-1-1'], '1-2'], '2', ['2-1', '2-2'], '3'] Any tips how to do that in Python? 回答1: def nested(flat, level=0): for k, it in itertools.groupby(flat, lambda x: x.split("-")[level]): yield next(it) remainder = list(nested(it, level + 1)) if remainder: yield