You can use DataFrame.select_dtypes to select string columns and then apply function str.strip. Notice: Values cannot be types like dicts or lists, because their dtypes is object. df_obj = df.select_dtypes([‘object’]) print (df_obj) 0 a 1 c df[df_obj.columns] = df_obj.apply(lambda x: x.str.strip()) print (df) 0 1 0 a 10 1 c 5 But if there are … Read more
You want \s Matches a single white space character, including space, tab, form feed, line feed. Equivalent to [ \f\n\r\t\v\u00a0\u1680\u2000-\u200a\u2028\u2029\u202f\u205f\u3000\ufeff] in Firefox and [ \f\n\r\t\v] in IE. str = str.replace(/\s/g, “X”);
Try List<string> parts = line.Split(‘;’).Select(p => p.Trim()).ToList(); FYI, the Foreach method takes an Action (takes T and returns void) for parameter, and your lambda return a string as string.Trim return a string Foreach extension method is meant to modify the state of objects within the collection. As string are immutable, this would have no effect … Read more
For space-character removal use “hello world”.replace(/\s/g, “”); for all white space use the suggestion by Rocket in the comments below!
You can use String#replaceAll() with a pattern of ^\”|\”$ for this. E.g. string = string.replaceAll(“^\”|\”$”, “”); To learn more about regular expressions, have al ook at http://regular-expression.info. That said, this smells a bit like that you’re trying to invent a CSV parser. If so, I’d suggest to look around for existing libraries, such as OpenCSV.
In short: strip() is “Unicode-aware” evolution of trim(). Meaning trim() removes only characters <= U+0020 (space); strip() removes all Unicode whitespace characters (but not all control characters, such as \0) CSR : JDK-8200378 Problem String::trim has existed from early days of Java when Unicode had not fully evolved to the standard we widely use today. … Read more
array_map() is what you need: $result = array_map(‘trim’, $source_array);
If you can modify the string: // Note: This function returns a pointer to a substring of the original string. // If the given string was allocated dynamically, the caller must not overwrite // that pointer with the returned value, since the original pointer must be // deallocated using the same allocator with which it … Read more
One line is enough: var x = ‘|f|oo||’; var y = x.replace(/^\|+|\|+$/g, ”); document.write(x + ‘<br />’ + y); ^ beginning of the string \|+ pipe, one or more times | or \|+ pipe, one or more times $ end of the string A general solution: function trim (s, c) { if (c === “]”) … Read more
Use this: if(typeof(String.prototype.trim) === “undefined”) { String.prototype.trim = function() { return String(this).replace(/^\s+|\s+$/g, ”); }; } The trim function will now be available as a first-class function on your strings. For example: ” dog”.trim() === “dog” //true EDIT: Took J-P’s suggestion to combine the regex patterns into one. Also added the global modifier per Christoph’s suggestion. … Read more