This question already has an answer here:
For loop for Dictionary don't follow it's order in Swift [duplicate]
(1 answer)
Closed 5 years ago.
the question that I want to ask, from which directions swift starts to read dictionaries or arrays
when I put some codes like this
let interestingNumber = [
"Square": [1,4,9,16,25],
"Prime": [2,3,5,7,11,13],
"Fiboannci": [1,1,2,3,5,8],
"asd":[2,3,4,5],
"zxc":[3,4,5]
]
for (key,values) in interestingNumber{
print(values)
}
the output is
[1, 4, 9, 16, 25]
[2, 3, 5, 7, 11, 13]
[1, 1, 2, 3, 5, 8]
[3, 4, 5]
[2, 3, 4, 5]
this is not the exact order, so do you know why swift does this ? and it sometimes makes it different too!
I guessed may be it does it in string order, then I tried but I think it is not too, so why swift does do that ?
Just like for NSDictionary, Swift dictionaries are not ordered by key or value. The order will always be unspecified.
If you need the keys to be sorted, your only option is to have an array of ordered keys, and use the for loop over this array.
From apple documentation (https://developer.apple.com/library/content/documentation/Swift/Conceptual/Swift_Programming_Language/CollectionTypes.html#//apple_ref/doc/uid/TP40014097-CH8-ID105)
Dictionaries
A dictionary stores associations between keys of the same type and
values of the same type in a collection with no defined ordering. Each
value is associated with a unique key, which acts as an identifier for
that value within the dictionary. Unlike items in an array, items in a
dictionary do not have a specified order. You use a dictionary when
you need to look up values based on their identifier, in much the same
way that a real-world dictionary is used to look up the definition for
a particular word.
Related
This question already has answers here:
Swift - How to mutate a struct object when iterating over it
(6 answers)
Closed 9 months ago.
What works in Python doesn't want to work in Swift.
Given is an array with subarrays containing numbers. During iteration, as soon as the condition exists that the element in question contains a certain value at position 0, I want to change the other values in this same subarray.
var coordinatesList = [[0, 1, 1, 1], [1, 12, 17, 23], [2, 81, 29, 66], [3, 41, 20, 94]]
for i in coordinatesList {
if i[0] == 6 {
i[3] = 5
}
}
However, my intention fails, because I get the error message that the element i is a let.
An error message with which I can do nothing at all and do not know how to deal with it.
You can iterate over the indices of the outer array. Then you can read and write the inner arrays using the indices:
for index in coordinatesList.indices {
if coordinatesList[index][0] == 6 {
coordinatesList[index][3] = 5
}
}
I want to do multiple where query without effect data. I want to get data that include at least 1 data per array. Pseudo code
data =[1,3]
array1 = [1,2]
array2 = [3,4]
if(data.IsIntersect(array1) and data.IsIntersect(array2))
IsIntersect checks are there a intersection beetween arrays
I did so far
queryBuilder.andWhere(
'properties.id IN (:...sizeIds) AND properties.id IN (:...colorIds)',
{ sizeIds: [1, 2], colorIds: [3, 4] },
);
It returns empty because firstly checks properties for 'sizeIds' then it checks for 'colorIds'. For example
properties includes 1,3
check for sizeIds, returns 1
check for colorIds, return empty
How can I do that with typeORM?
How can properties.id be 1 and 3? And if it is, how could 1 or 3 be in both? You're asking for the impossible.
I assume you mean to ask for when properties.id is 1 or 3, because if it is [1,3] then you should use the postgres array syntax {1,3} & the ANY keyword (some variation on this: Check if value exists in Postgres array).
tldr, I think all you need is brackets and OR instead of AND:
queryBuilder.andWhere(
'(properties.id IN (:...sizeIds) OR properties.id IN (:...colorIds))',
{ sizeIds: [1, 2], colorIds: [3, 4] },
);
If properties.id is in fact an array, then please add the entity definition to your question. If you want to merge the rows where properties.id is in the list you will need a GROUP BY (https://orkhan.gitbook.io/typeorm/docs/select-query-builder).
I'm using IndexSet and I'm trying to access some indexes which at times are consecutive and at other times are not.
For example, my set may contain [1, 2, 3, 5, 6, 7, 13, 31]
I want to pull out of the set a range of 3...13, but am having difficulty with the syntax. I've learned how to use the function commands from Apple documentation, by using myIndexSet.sorted(). However, the Apple Documentation does not give an example of how to access a range of elements in the set. The Apple Documentation for accessing elements in the index set are the following:
subscript(Range<IndexSet.Index>)
I've tried a number of ways to write this but can't figure out how to do it right. Can someone show me how to access a range of elements in the set to create a new set? I've tried things such as:
let subset = subscript(Range: myLargerSet.3...13)
but it doesn't seem to work.
Thanks
What you're looking for is the intersection of your IndexSet ([1, 2, 3, 5, 6, 7, 13, 31]) with another IndexSet ([3, 4, ..., 12, 13]):
let yourIndexSet: IndexSet = [1, 2, 3, 5, 6, 7, 13, 31]
let desiredIndexRange = IndexSet(3...13)
let indicesOfInterest = yourIndexSet.intersection(desiredIndexRange)
print(indicesOfInterest.sorted()) // => [3, 5, 6, 7, 13]
One possible solution is to use a filter to create a new IndexSet.
let set = IndexSet(arrayLiteral: 1,2,3,5,6,7,13,31)
let subset = set.filteredIndexSet { (index) -> Bool in
index >= 3 && index <= 13
}
You can access a slice of your original set as follows:
let slice = indexSet[indexSet.indexRange(in: 3...13)]
slice accesses the existing elements in place, so creation of the slice is O(1)
This question already has answers here:
Error: "array index out of range" in multidimensional array
(2 answers)
Closed 6 years ago.
I'm trying to populate a multidimensional array with this code:
var array = [[Int]]()
for i in 0...3 {
for j in 0...3{
array[i][j] = i + j <<- Error
}
}
But I get an error:
fatal error: Index out of range
What am I doing wrong?
[[Int]] is not a multidimensional array. It is an array of arrays. That's a very different thing. For example, in an array of arrays, each row may have a different number of columns. It's generally a bad idea to use a nested array as a multidimensional array, particularly a mutable one. It's often incredibly inefficient to modify because it causes a lot of copying every time you change it.
Swift doesn't have a multidimensional array type. If you really need one, you generally have to build it yourself, or redesign to avoid it. If it's small enough, and doesn't change much, it's not that big a deal, but don't let them get large.
That said, the problem is that element [0][0] doesn't exist because you didn't create it. You'd need to initialize the array this way before using it:
var array = Array(repeating: Array(repeating: 0, count: 4), count: 4)
This creates an array of 4 arrays of 4 zeros.
If you want specifically the layout you describe, possibly a better approach is mapping, which is likely going to be more efficient (since it doesn't keep modifying the nested array):
let array = (0...3).map { i in
(0...3).map { j in
return i + j
}
}
Calling array[i][j] is for elements that are already there. You cannot use it to initialize the array, because currently it is just an empty array. You should be using .append instead. Keep in mind that this actually isn't a multi-dimensional array like Rob Napier states, but it accomplishes the same goal in this scenario. Try something like this:
var array = [[Int]]()
for i in 0...3 {
var subArray = [Int]()
for j in 0...3 {
subArray.append(i + j)
}
array.append(subArray)
}
This prints:
[[0, 1, 2, 3], [1, 2, 3, 4], [2, 3, 4, 5], [3, 4, 5, 6]]
Again, may not be the best approach, but this is just how you could do it in Swift.
This question already has answers here:
NumPy selecting specific column index per row by using a list of indexes
(7 answers)
Closed 2 years ago.
Is there a better way to get the "output_array" from the "input_array" and "select_id" ?
Can we get rid of range( input_array.shape[0] ) ?
>>> input_array = numpy.array( [ [3,14], [12, 5], [75, 50] ] )
>>> select_id = [0, 1, 1]
>>> print input_array
[[ 3 14]
[12 5]
[75 50]]
>>> output_array = input_array[ range( input_array.shape[0] ), select_id ]
>>> print output_array
[ 3 5 50]
You can choose from given array using numpy.choose which constructs an array from an index array (in your case select_id) and a set of arrays (in your case input_array) to choose from. However you may first need to transpose input_array to match dimensions. The following shows a small example:
In [101]: input_array
Out[101]:
array([[ 3, 14],
[12, 5],
[75, 50]])
In [102]: input_array.shape
Out[102]: (3, 2)
In [103]: select_id
Out[103]: [0, 1, 1]
In [104]: output_array = np.choose(select_id, input_array.T)
In [105]: output_array
Out[105]: array([ 3, 5, 50])
(because I can't post this as a comment on the accepted answer)
Note that numpy.choose only works if you have 32 or fewer choices (in this case, the dimension of your array along which you're indexing must be of size 32 or smaller). Additionally, the documentation for numpy.choose says
To reduce the chance of misinterpretation, even though the following "abuse" is nominally supported, choices should neither be, nor be thought of as, a single array, i.e., the outermost sequence-like container should be either a list or a tuple.
The OP asks:
Is there a better way to get the output_array from the input_array and select_id?
I would say, the way you originally suggested seems the best out of those presented here. It is easy to understand, scales to large arrays, and is efficient.
Can we get rid of range(input_array.shape[0])?
Yes, as shown by other answers, but the accepted one doesn't work in general so well as what the OP already suggests doing.
I think enumerate is handy.
[input_array[enum, item] for enum, item in enumerate(select_id)]
How about:
[input_array[x,y] for x,y in zip(range(len(input_array[:,0])),select_id)]