dimensions

React Native Responsive Font Size

由 匿名 (未验证) 提交于 2019-12-03 01:23:02
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I would like to ask how react native handle or do the responsive font. For example in iphone 4s i Have fontSize: 14, while in iphone 6 I have fontSize: 18. 回答1: You can use PixelRatio for example: var React = require('react-native'); var {StyleSheet, PixelRatio} = React; var FONT_BACK_LABEL = 18; if (PixelRatio.get() Edit: Another example: import { Dimensions, Platform, PixelRatio } from 'react-native'; const { width: SCREEN_WIDTH, height: SCREEN_HEIGHT, } = Dimensions.get('window'); // based on iphone 5s's scale const scale = SCREEN_WIDTH /

how to obtain a CCSprite's width and height in cocos2d for iphone

由 给你一囗甜甜゛ 提交于 2019-12-02 20:16:42
That's the question xD Given an instance of a CCSprite in cocos2d in iphone, what method can I use to obtain the image width and height? The CCSprite class has a bounding box property that's a CGRect: CCSprite *sprite = [CCSprite spriteWithFile: @"file.png"]; int width = [sprite boundingBox].size.width; I added a width and height methods to my CCSprite subclass. -(CGFloat) width { return [self boundingBox].size.width; } -(CGFloat) height { return [self boundingBox].size.height; } yubenyi raw width: sprite.contentSize.width raw height: sprite.contentSize.height current width: sprite.contentSize

How to add a dimension to a numpy array in Python

由 不想你离开。 提交于 2019-12-02 11:22:57
I have an array that is size (214, 144). I need it to be (214,144,1) is there a way to do this easily in Python? Basically the dimensions are supposed to be (Days, Times, Stations). Since I only have 1 station's data that dimension would be a 1. However if I could also make the code flexible enough work for say 2 stations that would be great (e.g. changing the dimension size from (428,288) to (214,144,2)) that would be great! You could use reshape : >>> a = numpy.array([[1,2,3,4,5,6],[7,8,9,10,11,12]]) >>> a.shape (2, 6) >>> a.reshape((2, 6, 1)) array([[[ 1], [ 2], [ 3], [ 4], [ 5], [ 6]], [[

Firefox / IE textarea sizing quirk - workarounds?

由 老子叫甜甜 提交于 2019-12-02 11:13:09
Try out this code in Chrome, Firefox and IE: <!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html> <head> <title>Textarea problem</title> <style type="text/css"> html, body { position: absolute; top: 0; left: 0; right: 0; bottom: 0; border: 0; padding: 0; margin: 0; } #container { position: absolute; top: 4px;bottom: 4px; right: 4px;left: 4px; background-color: grey; } #ta { position: absolute; bottom: 0; right: 0; top: 0; left: 0; /*width: 100%; height: 100%;*/ border: black 4px solid; padding: 0; margin: 0; background-color: orange;

2019.10.26

由 ≡放荡痞女 提交于 2019-12-02 11:05:11
dimensions = (200,50) print(“Original dimensions:”) for dimension in dimensions: print(dimension) dimensions = (400,100) print(’\nModified dimensions:’) for dimension in dimensions: print(dimension) #小结元组是（ ），里面的元素不能修改，只能重新制定。 #【 】是列表，里面的元素可以修改 来源： https://blog.csdn.net/weixin_45707364/article/details/102754429

How can I solve this MATLAB “Matrix dimensions must agree” error?

由 早过忘川 提交于 2019-12-02 07:28:55
I am typing some code for a class but every time I run the function I get the same error: ??? Error using ==> plus Matrix dimensions must agree. Error in ==> Test at 6 f32=3.*exp((-x2.^2-y1.^2)./3); I know that the problem is a simple index error, but I can't seem to find it anywhere. Can somebody help me? Also I'm having the same problem with the following line of code: f34=(exp(-0.3./x2))./(log(y2).*sqrt(x2)); EDIT #1: x2 is defined as 0:0.1:5 and y1 is defined as -5:0.1:5 , but that is what I have been assigned to define them as. And I know exp is not a function because I have used it

How can I solve this MATLAB “Matrix dimensions must agree” error?

由 早过忘川 提交于 2019-12-02 06:56:11
问题 I am typing some code for a class but every time I run the function I get the same error: ??? Error using ==> plus Matrix dimensions must agree. Error in ==> Test at 6 f32=3.*exp((-x2.^2-y1.^2)./3); I know that the problem is a simple index error, but I can't seem to find it anywhere. Can somebody help me? Also I'm having the same problem with the following line of code: f34=(exp(-0.3./x2))./(log(y2).*sqrt(x2)); EDIT #1: x2 is defined as 0:0.1:5 and y1 is defined as -5:0.1:5 , but that is

resize image by area

由 岁酱吖の 提交于 2019-12-02 06:00:22
问题 I am trying to write a javascript function to resize an image based on a given area (or in my case (somewhat inaccurate) 'average dimension' since that's easier to think in terms of. Rather than feeding in maximum height and width, I want to feed in maximum area so that long or narrow images will appear visually to be roughly the same size. I'm getting really caught on the math aspect of it, though... just how to logic it, as I haven't done much math of late. Basically, given an aspect ratio

Android: Get Screen size for the root Layout only

由 佐手、 提交于 2019-12-02 05:18:54
问题 Please get me correctly over here : I want to get the height/width of the space available to the Activity/Layout in onCreate() method to calculate the height that can be given to child layouts. I can get the screen size using : root = (LinearLayout) findViewById(R.id.mainroot); // Main layout of LinearLayout android.view.Display display = getWindowManager().getDefaultDisplay(); int height = Display.getHeight(); // I know this is deprecated have hence used int width = Display.getWidth(); //

Firefox textarea sizing bug?

由 孤街浪徒 提交于 2019-12-02 00:05:36
问题 I'm attempting to create a fixed textarea that fills the entire width and height of the browser window using just CSS, and the following works perfectly in Chrome: textarea { border: 2px solid #3D6AA2; padding: 5px; position: fixed; top:0; left:0; right:0; bottom:0; resize: none; } http://jsfiddle.net/BulletzBill/FZr9k/1/ However, if you view the fiddle in Firefox, it appears as though it does not take the bottom or right rules into account at all. Is this a known bug in firefox, or is there
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