iguana | universal serialization engine | Serialization library
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kandi X-RAY | iguana Summary
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QUESTION
For example:
...ANSWER
Answered 2021-May-19 at 16:21You probably don't want to be using the actual characters making up the variable names in the code. As such, you'll need to explicitly define the category of each list:
QUESTION
For example:
...ANSWER
Answered 2021-May-18 at 23:39You can flatten the list of lists and then compare the elements in list1 with the flattened one using simple list comprehension.
QUESTION
I would like to determine whether a list of strings can be found within another list of strings in python.
For example:
...ANSWER
Answered 2021-May-18 at 21:53Your question was specifically about lists. I am presuming you are trying to ascertain if all elements in list2
are in list1
. If so, then we're talking about sets, and then this would be a solution:
QUESTION
so i'm fairly new to using python and I wanted to make a list of popular pets and have the person type in a number (that's assigned to an animal from a list of 10 in a dictionary) and print the correct animal from the number they typed in.
so far this is my code:
...ANSWER
Answered 2021-May-09 at 15:58Ok, two problems you have here.
First of all, a dictionary is like a door. You need a key to open a door. What is on the other side of the door is your value.
In your case, you want to access an animal, based on a number. The number will be a key and the animal will be your value.
So first step would be to reverse your dictionary key and values.
The second would be to change how you take an input and store it in a variable. Not like
QUESTION
I have a nested list that I am trying to split into smaller lists nested inside the original. I am splitting the lists by a specific element. I've figured that part out, but when I try to iterate through the original list, but only get the last list returned.
...ANSWER
Answered 2020-Dec-26 at 13:35animals = [["animal", "cat", "dog", "bird", "animal", "cat", "snake", "bird"], ["animal", "cat", "iguana", "bird", "animal", "lizard", "dog", "bird"]]
result = []
for row in animals:
result_row = []
for element in row:
if element == "animal":
if result_row:
result.append(result_row)
result_row = []
result_row.append(element)
result.append(result_row)
print(result)
QUESTION
package com.company.Iguana;
import javax.swing.event.ChangeEvent;
import javax.swing.event.ChangeListener;
import javax.swing.*;
import javax.swing.border.Border;
import java.awt.*;
import java.awt.event.InputEvent;
public class MyCPS implements ChangeListener {
JSlider Slider;
JPanel panel;
JLabel label;
JFrame frame;
JComboBox Combo;
MyCPS() {
frame = new JFrame("CPS");
panel = new JPanel();
label = new JLabel();
Slider = new JSlider();
String[] Buttons = {"Left Click", "Right Click"};
JComboBox combo = new JComboBox(Buttons);
ImageIcon image = new ImageIcon("BOAROR.png");
frame.setIconImage(image.getImage());
frame.getContentPane().setBackground(Color.YELLOW);
frame.setTitle("Iguana.exe");
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
frame.setSize(420, 420);
Slider = new JSlider(0, 23, 12);
Slider.setPreferredSize(new Dimension(400, 200));
Slider.setPaintTrack(true);
Slider.setMajorTickSpacing(4);
Slider.setPaintLabels(true);
Slider.setOrientation(SwingConstants.VERTICAL);
label.setText("CPS" + Slider.getValue());
label.setIcon(image);
Slider.addChangeListener(this);
panel.add(combo);
panel.add(Slider);
panel.add(label);
frame.add(panel);
frame.setSize(420,420);
frame.setVisible(true);
frame.pack();
}
@Override
public void stateChanged(ChangeEvent e) {
label.setText("CPS" + Slider.getValue());
if(e.equals(Slider.getValue())); <<<<<<<< Problem
System.out.println("Five");
}
...ANSWER
Answered 2020-Nov-27 at 23:25The error is because you are trying to compare the object, e
of type, ChangeEvent
with Slider.getValue()
which returns an int
.
You should write it as
QUESTION
Below is a sample of my dataframe.
...ANSWER
Answered 2020-Oct-25 at 04:37The following code can be used to support multiple aggregation functions for column R.
QUESTION
I am interested in reducing the time complexity for this code below with help of dictionary (or possibly another data structure).
As far as I understand my brute force solution has time complexity of O(n^2)
, however, possibly could be done in O(n)
(in n-times of non-nested loop).
The task is to print for for each day and location, the percentage of observations on that day and location that are mammal observations.
...ANSWER
Answered 2020-Jun-14 at 13:32There are a lot of really insane tricks to reduce time complexity but most of the time working with data you have to rely on how data ordered initialy. If it's not ordered that we can order it by some composite key. In your case the key is Tuple
where Item1 is day of date and Item2 is location. That would take n*log(n), and then use linear time to walk through the data producing results using linear time.
So looking at your data its already sorted. So we can skip that part and just walk through. Basic idea we initialize some state and chek if it changes we produce result. In our case state is current day, current location, and we track information in two variables total animals, total mammals.
QUESTION
So guys, I don't how to give a topic to this one. I'm having trouble to copy data from array of Object to another newly created array.
For example, I want to copy and create a new array
that contain all category of animals of every person in my database.
ANSWER
Answered 2020-May-29 at 04:31You could try using a Set. Set are made up of only unique values.
It wont matter if you push the same value it just won't add it to the set.
If you want to then convert the categories back to an array use the Array.from Method and pass in the set.
QUESTION
I have 3 models in Django :
...ANSWER
Answered 2020-May-21 at 12:56I would go rather with:
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