new Array(n) will create an array of n * undefined, so why can't you use new Array(n).map((_, index) => index) function to get [0, 1, 2 ..., n-1 ] What about arrays like this?
I know that new Array(n).fill(0).map((_, index) => index) is ok, but is there any essential difference between the two arrays of n * undefined and n * 0 ?
Common pits for initializing an n*m two-dimensional array:
Wrong way 1: new Array(n).fill(new Array(m).fill(0)) All arrays point to the same reference
Wrong way 2: new Array(n).map(v => new Array(m).fill(0)) "ghost array" problem
Correct way 1: new Array(n).fill(0).map(v => new Array(m).fill(0))
From the MDN Web Docs:
If the only argument passed to the Array constructor is an integer [...] this returns a new JavaScript array with its length property set to that number (Note: this implies an array of arrayLength empty slots, not slots with actual undefined values)
So like they say, it just sets the length, it doesn't set any of the values. So if the array contains no elements, then you can't map a non-existent value into a function.
Array Document
If the only argument passed to the Array constructor is an integer between 0 and 2^32 - 1 (inclusive), this returns a new JavaScript array with its length property set to that number (Note: this implies an array of arrayLength empty slots, not slots with actual undefined values — see sparse arrays). If the argument is any other number, a RangeError exception is thrown.
Array(n) or new Array(n) produces arrays with only length and no elements, so I call them "ghost arrays". Such an array cannot be traversed correctly with forEach or map (because there are no elements), but it is amazing that it has [##iterator], which can be traversed using for ... of, or expanded into an array using the spread operator (eg [...Array(10)]), and can be converted using Array.from().
Related
This strange behavior baffles me. I intend to create a dictionary with a single array field. Then within this array, two extra sub dictionaries are appended. Here is code,
var dictionary = [String: Any]()
var array = [[String: Any]]()
dictionary["array"] = array
var dict1:[String:Any] = ["abc": 123, "def": true]
var dict2:[String:Any] = ["111": 1.2345, "222": "hello"]
array.append(dict1)
array.append(dict2)
Debugger output.
As you can see from the debugger output, the var array is updated successfully (with 2 sub dictionaries appended). But the dictionary["array"] still has 0 value.
It appears the (dictionary["array"]) and (array) are two separate objects
Yes, they are separate. The element dictionary["array"] is an immutable value of type Array<_> because it's added as a value type to the dictionary not a reference type.
If you tried to add dict1 to the array by addressing the element via it's encapsulating dictionary like this:
(dictionary["array"] as! Array).append(dict1)
You would see an error like this:
error: cannot use mutating member on immutable value of type 'Array<_>'
From the Swift Language docs, emphasis added:
A value type is a type whose value is copied when it is assigned to a variable or constant, or when it is passed to a function.
You’ve actually been using value types extensively throughout the previous chapters. In fact, all of the basic types in Swift—integers, floating-point numbers, Booleans, strings, arrays and dictionaries—are value types, and are implemented as structures behind the scenes.
I need to find the index value of the object in an object list where the parameter x equals a certain value.
Is there a simple way to do this? I understand the IndexOf function, but how can I apply it to the object parameter?
You can use List.FindIndex:
int index = list.FindIndex(obj => obj.X == value);
The zero-based index of the first occurrence of an element that
matches the conditions defined by match, if found; otherwise, –1.
Functions can only return a single value but sometimes, we may want functions to return multiple values. Tuples can come in handy in such cases. We can create a tuple containing multiple values and return the tuple instead of a single value.
Write a function max_and_min that accepts a tuple containing integer elements as an argument and returns the largest and smallest integer within the tuple. The return value should be a tuple containing the largest and smallest value, in that order.
for example: max_and_min((1, 2, 3, 4, 5)) = (5, 1)
I am told to use an iteration to loop through each value of the tuple parameter to find the maximum and minimum values. Also, I must use Python 3.x.
How do I do this? I am really clueless. Thanks for your help!
def max_and_min(values):
# Write your code here
You are looking to pass a variable number of arguments to a function. In python, you can get multiple arguments passed at invocation with the * notation:
def max_and_min(*arg):
return (max(arg), min(arg))
Note that the Python 3 min and max functions themselves accept a variable number of arguments.
So I was getting a notice in my php while creating a google product feed.
The notice was
"The following php notice has occurred 4989 times on the _ site today:
PHP Notice: Undefined index: 0 in /xxx/Status.php on line 583"
This was the code in that class
public function inStockLocally($productcode)
{
if($this->_status[$productcode]['status'] == self::IN_STOCK) {
return $this->_status[$productcode]['in_stock_local'];
}
return false;
}
The function was getting a $productcode = 0, but the productcode was infact 'w32', so the key didn't exist.
up the stack where the function was being called I put this in, in order to break on the troublesome product.
if ($productcode == 0) {
$test = 'breakhere';
}
Using netbeans and firebug, it broke on the line when $productcode = 'w32'
So my question is why does 'w32' == 0 evaluate to true?
It is also evaluating to true with other similar structure codes like 'h94'.
Any help would be appreciated as no one in the department can figure out why this is happening.
I guess I didn't put enough info in the q. Two things going on.
1. 'w32' converted to a number = 0 for some reason. 2. [0] is being inserted as my key in the array when the productcode has the structure 'x##';
I'm a little new here, so pardon if this isn't the answer you were expecting, but PHP does a lot of automatic type conversion. So any string that doesn't start with a numeric character (0..9, +, -, etc) will evaluate to zero.
"If you compare a number with a string or the comparison involves numerical strings, then each string is converted to a number and the comparison performed numerically. "
http://php.net/manual/en/language.operators.comparison.php
Additionally, I suppose you have an indexed array, although you expect it to be an associative array:
The array() function is used to create an array.
In PHP, there are three types of arrays:
Indexed arrays - Arrays with numeric index
Associative arrays - Arrays with named keys
Multidimensional arrays - Arrays containing one or more arrays
Syntax
Syntax for indexed arrays:
array(value1,value2,value3,etc.);
Syntax for associative arrays:
array(key=>value,key=>value,key=>value,etc.);
I came across this bit of code:
n = args[0] as Long
[*n..1, n].any{ println ' '*it + '*'*(n - ~n - it*2) }
It's used for printing a tree form of structure. Like this:
*
***
*****
*******
*
(for n=4)
How does the code [*n..1,n] produce [4, 3, 2, 1, 4]?
How does any method works here? The Doc doesn't help me much. What is a predictive that can be passed to any(as mentioned in Doc's)?
Whats the use of any and how its handled in this case?
Q1a: * "unpacks" an array. .. creates a range. [] creates a collection.
Q1b: *n..1 unpacks [4,3,2,1] into its individual parts.
Q1c: [4,3,2,1,n] == [4,3,2,1,4]
Q2: I don't know why any was used here; each works just as well, and makes more sense in context. any does loop over the connection, so the println side-effect functions as intended.
Normally any would be used to determine if any collection elements met a criteria, for example:
[*n..1,n].any { it > 10 } // Returns false, no elements are > 10
[*n..1,n].any { it == 3 } // Returns true, because at least one element is 3
The last statement of the closure is used to determine if each item meets the criteria. println returns null, so any will return false. The value is unused and discarded.
The only reason I can think of that someone might have used any is to avoid seeing the return value of each in the console. each returns the original collection.
1) n..1 is called a range literal, it creates a groovy.lang.Range object that decrements by 1 from n to 1. This is then merged into the surrounding list context using the "Spread operator (*)"
2) the any method is defined in DefaultGroovyMethods and it is a predicate function that returns true if an element in a collection satisfies the supplied predicate closure. In this example, the code doesn't check the return value, so original other could have produced the same output using an each call instead.