less 使用二维数组

时间: 2023-11-27 22:36:13 浏览: 31
在 Less 中并没有直接支持二维数组的语法,因为 Less 是一种 CSS 预处理器,主要用于增强 CSS 的功能,而 CSS 本身并不支持多维数组。 不过你可以使用嵌套的列表(nested list)来模拟二维数组的效果。嵌套的列表是 Less 中的一种数据结构,类似于多维数组。 以下是一个示例: ```less @twoDArray: (1, 2, 3), (4, 5, 6), (7, 8, 9); @value: extract(@twoDArray, 2, 3); // 获取第三行第四列的值 .my-element { width: extract(extract(@twoDArray, 2), 3); // 获取第三行第四列的值,并应用于元素的宽度 } ``` 在上述示例中,我们使用了嵌套的列表 `@twoDArray` 来表示一个二维数组。然后,可以使用 `extract()` 函数来获取特定位置的值。在上述示例中,我们获取了第三行第四列的值,并将其应用于 `.my-element` 元素的宽度样式。 请注意,嵌套列表在 Less 中仅用作数据结构,而不是用于进行类似于编程语言中的数组操作。因此,它们的功能有限,并且不能像二维数组那样进行复杂的操作。
相关问题

使用numpy库判断二维数组中是否有负数

可以使用numpy库中的函数np.any()和np.less()来判断二维数组中是否有负数。具体实现代码如下: import numpy as np arr = np.array([[1, 2, 3], [4, -5, 6], [7, 8, 9]]) if np.any(np.less(arr, )): print("二维数组中存在负数") else: print("二维数组中不存在负数")

设计算法并编写程序,定义、输入10行10列二维数组,分别求其中大于等于60和小于60的元素个数

下面是一个Python实现的例子: ```python array = [] # 定义二维数组 # 输入10行10列的二维数组 for i in range(10): row = input().split() row = [int(x) for x in row] array.append(row) # 初始化计数器 count_greater_than_or_equal_to_60 = 0 count_less_than_60 = 0 # 遍历二维数组并统计个数 for row in array: for element in row: if element >= 60: count_greater_than_or_equal_to_60 += 1 else: count_less_than_60 += 1 # 输出结果 print("大于等于60的元素个数:", count_greater_than_or_equal_to_60) print("小于60的元素个数:", count_less_than_60) ``` 这里我们使用了两个计数器,分别统计大于等于60和小于60的元素个数。在遍历二维数组时,我们用if语句判断当前元素的大小并根据情况更新计数器。最后输出两个计数器即可。

相关推荐

用c++解决Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operations only with these currencies. There can be several points specializing in the same pair of currencies. Each point has its own exchange rates, exchange rate of A to B is the quantity of B you get for 1A. Also each exchange point has some commission, the sum you have to pay for your exchange operation. Commission is always collected in source currency. For example, if you want to exchange 100 US Dollars into Russian Rubles at the exchange point, where the exchange rate is 29.75, and the commission is 0.39 you will get (100 - 0.39) * 29.75 = 2963.3975RUR. You surely know that there are N different currencies you can deal with in our city. Let us assign unique integer number from 1 to N to each currency. Then each exchange point can be described with 6 numbers: integer A and B - numbers of currencies it exchanges, and real RAB, CAB, RBA and CBA - exchange rates and commissions when exchanging A to B and B to A respectively. Nick has some money in currency S and wonders if he can somehow, after some exchange operations, increase his capital. Of course, he wants to have his money in currency S in the end. Help him to answer this difficult question. Nick must always have non-negative sum of money while making his operations. Input The first line contains four numbers: N - the number of currencies, M - the number of exchange points, S - the number of currency Nick has and V - the quantity of currency units he has. The following M lines contain 6 numbers each - the description of the corresponding exchange point - in specified above order. Numbers are separated by one or more spaces. 1 ≤ S ≤ N ≤ 100, 1 ≤ M ≤ 100, V is real number, 0 ≤ V ≤ 103. For each point exchange rates and commissions are real, given with at most two digits after the decimal point, 10-2 ≤ rate ≤ 102, 0 ≤ commission ≤ 102. Let us call some sequence of the exchange operations simple if no exchange point is used more than once in this sequence. You may assume that ratio of the numeric values of the sums at the end and at the beginning of any simple sequence of the exchange operations will be less than 104. Output If Nick can increase his wealth, output YES, in other case output NO.

Dog Card is a card game. In the game, there are a total of 2n cards in the deck, each card has a value, and the values of these 2n cards form a permutation of 1 ~ 2n. There is a skill that works as follows: 1. Draw a card from the top of the deck. 2. If the deck is empty, then skip to step 3, otherwise you guess whether the card on the top of the deck has a higher value than your last drawn card and draw a card from the top of the deck. If your guess is correct, then repeat this step, otherwise skip to step 3. 3. End this process. Nana enjoys playing this game, although she may not be skilled at it. Therefore, her guessing strategy when using this skill is simple: if the value of the last drawn card is less than or equal to n, then she guesses that the next oard's valve is higher, ther wse, she guedses thet the next card's vaue s lomler she wârns tb dmokt tor anfafrhlm decks of cards (Obviously, there are (2n)! cases), how many cards she can draw in total if she uses the skill only once in each case. Since this number can be very large,please provide the answer modulo a given value.Dog Card is a card game. In the game, there are a total of 2n cards in the deck, each card has a value, and the values of these 2n cards form a permutation of 1 ~ 2n. There is a skill that works as follows: 1. Draw a card from the top of the deck. 2. If the deck is empty, then skip to step 3, otherwise you guess whether the card on the top of the deck has a higher value than your last drawn card and draw a card from the top of the deck. If your guess is correct, then repeat this step, otherwise skip to step 3. 3. End this process. Nana enjoys playing this game, although she may not be skilled at it. Therefore, her guessing strategy when using this skill is simple: if the value of the last drawn card is less than or equal to n, then she guesses that the next oard's valve is higher, ther wse, she guedses thet the next card's vaue s lomler she wârns tb dmokt tor anfafrhlm decks of cards (Obviously, there are (2n)! cases), how many cards she can draw in total if she uses the skill only once in each case. Since this number can be very large,please provide the answer modulo a given value.给出c++代码及中文解释

最新推荐

recommend-type

pre_o_1csdn63m9a1bs0e1rr51niuu33e.a

pre_o_1csdn63m9a1bs0e1rr51niuu33e.a
recommend-type

matlab建立计算力学课程的笔记和文件.zip

matlab建立计算力学课程的笔记和文件.zip
recommend-type

FT-Prog-v3.12.38.643-FTD USB 工作模式设定及eprom读写

FT_Prog_v3.12.38.643--FTD USB 工作模式设定及eprom读写
recommend-type

matlab基于RRT和人工势场法混合算法的路径规划.zip

matlab基于RRT和人工势场法混合算法的路径规划.zip
recommend-type

matlab基于matlab的两步定位软件定义接收机的开源GNSS直接位置估计插件模块.zip

matlab基于matlab的两步定位软件定义接收机的开源GNSS直接位置估计插件模块.zip
recommend-type

zigbee-cluster-library-specification

最新的zigbee-cluster-library-specification说明文档。
recommend-type

管理建模和仿真的文件

管理Boualem Benatallah引用此版本:布阿利姆·贝纳塔拉。管理建模和仿真。约瑟夫-傅立叶大学-格勒诺布尔第一大学,1996年。法语。NNT:电话:00345357HAL ID:电话:00345357https://theses.hal.science/tel-003453572008年12月9日提交HAL是一个多学科的开放存取档案馆,用于存放和传播科学研究论文,无论它们是否被公开。论文可以来自法国或国外的教学和研究机构,也可以来自公共或私人研究中心。L’archive ouverte pluridisciplinaire
recommend-type

实现实时数据湖架构:Kafka与Hive集成

![实现实时数据湖架构:Kafka与Hive集成](https://img-blog.csdnimg.cn/img_convert/10eb2e6972b3b6086286fc64c0b3ee41.jpeg) # 1. 实时数据湖架构概述** 实时数据湖是一种现代数据管理架构,它允许企业以低延迟的方式收集、存储和处理大量数据。与传统数据仓库不同,实时数据湖不依赖于预先定义的模式,而是采用灵活的架构,可以处理各种数据类型和格式。这种架构为企业提供了以下优势: - **实时洞察:**实时数据湖允许企业访问最新的数据,从而做出更明智的决策。 - **数据民主化:**实时数据湖使各种利益相关者都可
recommend-type

2. 通过python绘制y=e-xsin(2πx)图像

可以使用matplotlib库来绘制这个函数的图像。以下是一段示例代码: ```python import numpy as np import matplotlib.pyplot as plt def func(x): return np.exp(-x) * np.sin(2 * np.pi * x) x = np.linspace(0, 5, 500) y = func(x) plt.plot(x, y) plt.xlabel('x') plt.ylabel('y') plt.title('y = e^{-x} sin(2πx)') plt.show() ``` 运行这段
recommend-type

JSBSim Reference Manual

JSBSim参考手册,其中包含JSBSim简介,JSBSim配置文件xml的编写语法,编程手册以及一些应用实例等。其中有部分内容还没有写完,估计有生之年很难看到完整版了,但是内容还是很有参考价值的。