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Override List(of String) Elements in a Sub


Override List(of String) Elements in a Sub

By : user2185489
Date : November 22 2020, 04:01 AM
wish of those help You have to use a For-loop, you can't modify the string with a For Each:
code :
Sub RemoveFirstChar(ByRef myList As List(Of String))
    For i As Integer = 0 To myList.Count - 1
        Dim str As String = myList(i)
        myList(i) = str.Substring(2)
    Next
End Sub
myList(i) = If(str.Length >= 1, str.Substring(1), str)


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Obtain a list containing string elements excluding elements prefixed with any other element from initial list

Obtain a list containing string elements excluding elements prefixed with any other element from initial list


By : user3626744
Date : March 29 2020, 07:55 AM
this one helps. This algorithm completes the task in 0.97 second on my computer, with the input file submitted by the author (154MB):
code :
l.sort()

last_str = l[0]
filtered = [last_str]
app      = filtered.append

for str in l:
    if not str.startswith(last_str):
        last_str = str
        app(str)

# Commented because of the massive amount of data to print.
# print filtered
How do I override the toString method to display elements in a list in Java?

How do I override the toString method to display elements in a list in Java?


By : Nam Nhi
Date : March 29 2020, 07:55 AM
wish of those help toString() has no arguments. Overwrite it like so (assuming you are extending a List class):
code :
@Override
public String toString() {
    String result = " ";
    for (int i = 0; i < this.size(); i++) {
        result += " " + this.get(i);
    }
    return result;
}
@Override
public String toString() {
    StringBuilder sb = new StringBuilder();
    sb.append(" "); // remove this if you do not want two spaces at the front of the returned string
    for (int i = 0; i < this.wordList.size(); i++) {
        sb.append(" " + this.wordList.get(i).toString());
    }
    return sb.toString();
}
public static String getStringRepresentation(List<String> list) {
    StringBuilder sb = new StringBuilder();
    sb.append(" "); // remove this if you do not want two spaces at the front of the returned string
    for (int i = 0; i < list.size(); i++) {
        sb.append(" " + list.get(i).toString());
    }
    return sb.toString();
}
public static void main(String[] args) {
    List<String> list = new ArrayList<String>();
    list.add("foo");
    list.add("bar");
    String listStringRepr = WordsContainer.getStringRepresentation(list);
    System.out.println(listStringRepr);
}
Search list of string elements that match another list of string elements

Search list of string elements that match another list of string elements


By : Carlos Guerrero
Date : March 29 2020, 07:55 AM
To fix this issue Use the any() function with a generator expression:
code :
a = [x for x in names if any(pat in x for pat in pattern)]
How to override cascading of CSS font-size back to default on child elements (in a list)

How to override cascading of CSS font-size back to default on child elements (in a list)


By : D Shop
Date : March 29 2020, 07:55 AM
To fix this issue Actually font-size: initial would work in Edge and many other browsers but that won't help you here.
The issue here is that the initial value works differently than you might expect - it resets everything to spec default. As a result, a h2 being rendered small is correct.
How to filter string elements in the list which occurs in longer elements in the same list python?

How to filter string elements in the list which occurs in longer elements in the same list python?


By : brentongillis
Date : March 29 2020, 07:55 AM
I wish this helpful for you This solution uses brute force to check each string against the remaining (slicing the array), justo for the fun of writing a one liner.
It does not removes string if the order is reversed, for example "hair bright" is not considered as sub of "bright hair dryer".
code :
[ e for i, e in enumerate(mylist) if not any([ e in s for s in mylist[:i] + mylist[i+1:] ]) ]

   #=> ['love', 'bright light', 'bright hair dryer']
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