You are misunderstanding that particular overload of the LastIndexOf method. The docs state the following: The search starts at a specified character position and proceeds backward toward the beginning of the string. Note that it says backward. So, if you start at position 0, there is no “<” substring at that position or in front … Read more
int index = employeeList.FindIndex(employee => employee.LastName.Equals(somename, StringComparison.Ordinal)); Edit: Without lambdas for C# 2.0 (the original doesn’t use LINQ or any .NET 3+ features, just the lambda syntax in C# 3.0): int index = employeeList.FindIndex( delegate(Employee employee) { return employee.LastName.Equals(somename, StringComparison.Ordinal); });
indexOf compares searchElement to elements of the Array using strict equality (the same method used by the ===, or triple-equals, operator). You cannot use === to check the equability of an object. As @RobG pointed out Note that by definition, two objects are never equal, even if they have exactly the same property names and … Read more
You have to use code like this. int index = s.IndexOf(‘,’, s.IndexOf(‘,’) + 1); You may need to make sure you do not go outside the bounds of the string though. I will leave that part up to you.
string s = “hello”; char c = s[1]; // now c == ‘e’ See also Substring, to return more than one character.
Use overloaded version of indexOf(), which takes the starting index (fromIndex) as 2nd parameter: str.indexOf(“is”, str.indexOf(“is”) + 1);
Use np.where to get the indices where a given condition is True. Examples: For a 2D np.ndarray called a: i, j = np.where(a == value) # when comparing arrays of integers i, j = np.where(np.isclose(a, value)) # when comparing floating-point arrays For a 1D array: i, = np.where(a == value) # integers i, = np.where(np.isclose(a, … Read more
You really could use the regular expression /((s).*?){n}/ to search for n-th occurrence of substring s. In C# it might look like this: public static class StringExtender { public static int NthIndexOf(this string target, string value, int n) { Match m = Regex.Match(target, “((” + Regex.Escape(value) + “).*?){” + n + “}”); if (m.Success) return … Read more
var str = “I learned to play the Ukulele in Lebanon.” var regex = /le/gi, result, indices = []; while ( (result = regex.exec(str)) ) { indices.push(result.index); } UPDATE I failed to spot in the original question that the search string needs to be a variable. I’ve written another version to deal with this case … Read more
Maybe you would like to use higher-order functions such as “map”. Assuming you want search by ‘field’ attribute: var elementPos = array.map(function(x) {return x.id; }).indexOf(idYourAreLookingFor); var objectFound = array[elementPos];