trim

Is there a Perl-compatible regular expression to trim whitespace from both sides of a string?

我只是一个虾纸丫 提交于 2019-12-03 16:59:10
问题 Is there a way to do this in one line? $x =~ s/^\s+//; $x =~ s/\s+$//; In other words, remove all leading and trailing whitespace from a string. 回答1: $x =~ s/^\s+|\s+$//g; or s/^\s+//, s/\s+$// for $x; 回答2: My first question is ... why? I don't see any of the single-regexp solutions to be any more readable than the regexp you started with. And they sure aren't anywhere near as fast. #!/usr/bin/perl use strict; use warnings; use Benchmark qw(:all); my $a = 'a' x 1_000; my @x = ( " $a ", "$a ",

java.lang.String的trim()方法详解

别等时光非礼了梦想. 提交于 2019-12-03 14:42:47
String.Trim()方法到底为我们做了什么,仅仅是去除字符串两端的空格吗? 一直以为Trim()方法就是把字符串两端的空格字符给删去,其实我错了,而且错的比较离谱。 首先我直接反编译String类,找到Trim()方法: public string Trim() { return this.TrimHelper(WhitespaceChars, 2); } TrimHelper方法有两个参数,第一个参数名WhitespaceChars,首字母尽然是大写的,肯定有文章,真不出我所料: internal static readonly char[] WhitespaceChars; 这里只是定义它,没有赋值,而且是静态的,我们看看构造函数去,果然找到: static String() { Empty = " "; WhitespaceChars = new char [] { '/t', '/n', '/v', '/f', '/r', ' ', '/x0085', '/x00a0', '?', ' ', ' ', ' ', ' ', '?', '?', '?', '?', '?', ' ', '?', '?', '/u2028', '/u2029', ' ', '?' }; } Trim方法就是把字符串两端的这些字符给删去?我很坚定的猜想到。 继续我们的探索

Trim string column in PySpark dataframe

风格不统一 提交于 2019-12-03 09:58:45
I'm beginner on Python and Spark. After creating a DataFrame from CSV file, I would like to know how I can trim a column. I've try: df = df.withColumn("Product", df.Product.strip()) df is my data frame, Product is a column in my table But I see always the error: Column object is not callable Do you have any suggestions? Starting from version 1.5 , Spark SQL provides two specific functions for trimming white space, ltrim and rtrim (search for "trim" in the DataFrame documentation ); you'll need to import pyspark.sql.functions first. Here is an example: from pyspark.sql import SQLContext from

Mysql:Trim all fields in database

半腔热情 提交于 2019-12-03 09:40:37
UPDATE mytable SET mycolumn= LTRIM(RTRIM(mycolumn)); works fine on trimming columns removing trailer spaces, but how can i adjust it to trim all columns without having to write each column name in table ?? cause i kind have a huge database. Some years late, but might help others: This code trims all fields of a the table "your_table". Could be expanded to work on the whole database in the same way.... SET SESSION group_concat_max_len = 1000000; select concat('update your_table set ',group_concat(concat('`',COLUMN_NAME, '` = trim(`',COLUMN_NAME,'`)')),';') FROM INFORMATION_SCHEMA.COLUMNS WHERE

Is there a Perl-compatible regular expression to trim whitespace from both sides of a string?

守給你的承諾、 提交于 2019-12-03 05:58:31
Is there a way to do this in one line? $x =~ s/^\s+//; $x =~ s/\s+$//; In other words, remove all leading and trailing whitespace from a string. $x =~ s/^\s+|\s+$//g; or s/^\s+//, s/\s+$// for $x; My first question is ... why? I don't see any of the single-regexp solutions to be any more readable than the regexp you started with. And they sure aren't anywhere near as fast. #!/usr/bin/perl use strict; use warnings; use Benchmark qw(:all); my $a = 'a' x 1_000; my @x = ( " $a ", "$a ", $a, " $a" ); cmpthese(-5, { single => sub { for my $s (@x) { my $x = $s; $x =~ s/^\s+|\s+$//g; } }, double =>

How to delete specific characters from a string in Ruby?

柔情痞子 提交于 2019-12-03 05:21:37
问题 I have several strings that look like this: "((String1))" They are all different lengths. How could I remove the parentheses from all these strings in a loop? 回答1: Do as below using String#tr : "((String1))".tr('()', '') # => "String1" 回答2: If you just want to remove the first two characters and the last two, then you can use negative indexes on the string: s = "((String1))" s = s[2...-2] p s # => "String1" If you want to remove all parentheses from the string you can use the delete method on

trim is not part of the standard c/c++ library?

筅森魡賤 提交于 2019-12-03 05:09:33
Is it me or are there no standard trim functions in the c or c++ library? is there any single function that acts as a trim? If not can anyone tell me Why trim is not part of the standard library? (i know trim is in boost) My trim code is std::string trim(const std::string &str) { size_t s = str.find_first_not_of(" \n\r\t"); size_t e = str.find_last_not_of (" \n\r\t"); if(( string::npos == s) || ( string::npos == e)) return ""; else return str.substr(s, e-s+1); } test: cout << trim(" \n\r\r\n \r\n text here\nwith return \n\r\r\n \r\n "); -edit- i mostly wanted to know why it wasnt in the

Incorrect decrement of the reference count of an object that is not owned at this point by the caller

假如想象 提交于 2019-12-03 03:57:16
Incorrect decrement of the reference count of an object that is not owned at this point by the caller on iPhone. It is happening with NSString which I clearly init and release within the for loop. I have tried to do the same as an autoreleases string but I get leaks. I assume the culprit is the stringbytrimming call. Any suggestions, by the way this does not leak, but I get the warning in build and analyze. Everything also works fine and the app does not crash. for(int i=0;i<storyQuantity;i++) { NSString *imageString = [[NSString alloc] init]; imageString = [[[storiesArray objectAtIndex:i]

Trim whitespace from a String

家住魔仙堡 提交于 2019-12-03 03:45:26
问题 I know there are several ways to do this in Java and C that are nice, but in C++ I can't seem to find a way to easily implement a string trimming function. This is what I currently have: string trim(string& str) { size_t first = str.find_first_not_of(' '); size_t last = str.find_last_not_of(' '); return str.substr(first, (last-first+1)); } but whenever I try and call trim(myString); I get the compiler error /tmp/ccZZKSEq.o: In function `song::Read(std::basic_ifstream<char, std::char_traits

How many spaces will Java String.trim() remove?

不问归期 提交于 2019-12-03 02:54:01
问题 In Java, I have a String like this: " content ". Will String.trim() remove all spaces on these sides or just one space on each? 回答1: All of them. Returns : A copy of this string with leading and trailing white space removed, or this string if it has no leading or trailing white space. ~ Quoted from Java 1.5.0 docs (But why didn't you just try it and see for yourself?) 回答2: From the source code (decompiled) : public String trim() { int i = this.count; int j = 0; int k = this.offset; char[]