digits

cv::Mat matrix, HOW TO Reduce digits to the right of the decimal point in cv::Mat?

半腔热情 提交于 2019-12-06 01:41:49
问题 I have an app that prints a 3x3 cv::Mat on the iPhone screen. I need to reduce the decimals, as the screen is not so big, see: [1.004596557012473, -0.003116992336797859, 5.936915104939593; -0.007241746117066327, 0.9973985665720294, -0.2118670500989478; 1.477734234970711e-05, -1.03363734495053e-05, 1.000089074805124] so I would like to reduce the decimals .4 or .6 or six decimals. Any ideas? Cheers 回答1: If you were using printf cv::Mat data(3, 3, CV_64FC1); for (int y = 0; y < data.rows; ++y)

How can you get the number of digits contained in a double?

风格不统一 提交于 2019-12-06 01:19:07
I'm trying to get the number of digits in the following double value: 56.46855976 without using converting it to a string (and simply replacing the "." with a ""). Anybody got any ideas? Count how often you must divide the number by 10 until it's smaller than 1 -> that gives you the digits before the decimal point. Then count how often you must multiply the original number by 10 until it equals the Math.Floor-result -> that gives you the digits behind the decimal points. Add. Be glad. Edit : As Joey points out, there is some uncertianity in it. Define a maximum number of digits beforehand so

Problem on getting digits in a double variable

倾然丶 夕夏残阳落幕 提交于 2019-12-05 13:03:30
I got a little problem with a function I need in my java progam. I want to check the total amount of digits, a 'double' variable has. (e.g.: 5 should return 1, 5.0034 should return 5, and 2.04 should return 3.) My function is this: private int getDoubleLength (double a) { int len = 0; while (a != (int) a) { a *= 10; } while (a > 1) { len ++; a/=10; } System.out.println(String.format("Double %f has %d digits.",a,len)); return len; } Now this works perfectly, but when I am making some mathematical calculations, you get the slight errors of doubles: I divide 10 by cos(60), which is 0,5. The

Inserting spaces between digits in C

风流意气都作罢 提交于 2019-12-05 10:43:25
How would I go about taking a number like 123456 and having it print as 1 2 3 4 5 6 ? As 'jamesdlin' has mentioned in his comment, GMan's approach would work, however you will need to store it in a buffer in order to print out in the correct order (his algorithm would print out "6 5 4 3 2 1" for input 123456). At this point I'd say that it would be much simpler to just use sprintf as 'therefromhere' suggested in his answer (if this is not an algorithm class assignment of course). In my opinion the simplest way to do it would be using recursion, this way you can print out digits in the right

do I have to specify integer length when creating an id field in MySQL through phpMyAdmin?

丶灬走出姿态 提交于 2019-12-05 09:19:54
I saw someone not set the length in a tutorial but it was specifically for counting the total number of users and just set to auto-increment. I've been of the habit of always specifying a length because I thought it was mandatory, but I wanted to ask if I can leave it blank unless it specifically a date or pin number etc where the length is always set. (I used to set it as 11 digits or more if I wasn't sure) Every integer field defaults to 11 when left blank so you can leave it. No, you don't have to specify a length for integers. I have never done that. There are different integral data types

Converting Arabic numerals to Arabic/Persian numbers in html file

元气小坏坏 提交于 2019-12-05 05:16:58
问题 I am trying to convert the plain text Arabic Numerals into Eastern Arabic digits. So basically taking 1 2 3... and converting them into ١‎ ٢‎ ٣‎... . The function converts all numbers, including any numbers contained within tags, i.e. H1 . private void LoadHtmlFile(object sender, EventArgs e) { var htmlfile = "<html><body><h1>i was born in 1988</h1></body></html>".ToArabicNumber(); ; webBrowser1.DocumentText=htmlfile; } } public static class StringHelper { public static string ToArabicNumber

One or two numeric digits Regex

我的未来我决定 提交于 2019-12-04 23:45:13
I have the below code. It works only when I have 2 digits. If I have 1 digit doesn't work. I want to work in both cases: one or two digit. var numberRegex = /^[1-9][0-9]$/; I've tried something like this but unfortunately doesn't work: var numberRegex = /^[1-9]?[1-9][0-9]$/; Thanks for support. Try this one out: /^\d{1,2}$/; Reading what you have it looks like you don't want to accept numbers like 01 . /^\d{1}|[1-9]\d{1}$/; Try this. /^[0-9]|[0-9][0-9]$/ This should do the job. Using an Or operator does it. try this regex: /^[1-9]\d{0,1}$/ Ozz This works: /^([0-9]{0,1}([1-9][0-9]){0,2})$/ 来源:

Set edittext using only “,0123456789” programmatically in Android?

允我心安 提交于 2019-12-04 23:29:17
My question is this. I can say in xml android:digits="0123456789," But this time I've to it add trough java code. customEditText.setKeyListener(DigitsKeyListener.getInstance("0123456789,")); Doesn't work. Application gives me same result as android:InputType="number" So is there any alternative way to say android:digits using java? EDITED So my question wasn't how to get android:InputType="number" using java. My question is that how I can get android:digits="0123456789," to be used with java. At this point I don't care if the user can see characters. I just want my edittext field to accept

C# string starts with a number regex

こ雲淡風輕ζ 提交于 2019-12-04 22:47:42
I've been searching around for a little while to figure out how to confirm a string begins with a number. I came across a regex statement '/^\d/' which I can confirm says if starts with digit. However I can not seem to get it to work in the code below. Where did I went wrong in the statement if(Regex.Match(info,"/^\d/")) ? //String attachbody is the attachment body from an email C read into a string string[] data = Regex.Split(attachbody, "\n"); foreach (String info in data) { if (Regex.Match(info,"/^\d/")) { string[] tabbedHeaderData = Regex.Split(info, "\t"); TicketID = tabbedHeaderData[0]

Why `(map digitToInt) . show` is so fast?

人盡茶涼 提交于 2019-12-04 18:41:31
问题 Converting non-negative Integer to its list of digits is commonly done like this: import Data.Char digits :: Integer -> [Int] digits = (map digitToInt) . show I was trying to find a more direct way to perform the task, without involving a string conversion, but I'm unable to come up with something faster. Things I've been trying so far: The baseline: digits :: Int -> [Int] digits = (map digitToInt) . show Got this one from another question on StackOverflow: digits2 :: Int -> [Int] digits2 =