Having a little trouble tracking down the Swift equivalent of:
//timeArray and locationArray are NSMutableArrays
NSRange removalRange = NSMakeRange(0, i);
[timeArray removeObjectsInRange:removalRange];
[locationArray removeObjectsInRange:removalRange];
I see that Swift does have a call in the API: typealias NSRange = _NSRange but I haven't got past that part. Any help?
In addition to Antonio's answer, you can also just use the range operator:
var array = [0, 1, 2, 3, 4, 5]
array.removeRange(1..<3)
// array is now [0, 3, 4, 5]
The half-closed range operator (1..<3) includes 1, up to but not including 3 (so 1-2).
A full range operator (1...3) includes 3 (so 1-3).
Use the removeRange method of the swift arrays, which requires an instance of the Range struct to define the range:
var array = [1, 2, 3, 4]
let range = Range(start: 0, end: 1)
array.removeRange(range)
This code removes all array elements from index 0 (inclusive) up to index 1 (not inclusive)
Swift 3
As suggested by #bitsand, the above code is deprecated. It can be replaced with:
let range = 0..<1
array.removeSubrange(range)
or, more concisely:
array.removeSubrange(0..<1)
Related
I'm currently trying to apply a mask to a column on a dataframe, in order to gain the mean from certain values. However, I don't want to do this over the whole column, just over a small range. This is my code at present:
data = pd.DataFrame({"test":[12, 4, 5, 4, 1, 3, 2, 5, 10, 9, 4, 3, 2, 1, 4, 2, 2, 4, 2, 5]})
range_start = 5
range_finish = 17
mask = np.arange(len(data)) %4
measured_stress_ratio_overload = data.iloc[range_start:range_finish, mask == 0, 'test'].mean()
measured_stress_ratio_baseline = data.iloc[range_start:range_finish, mask!= 0, 'test'].mean()
My expected output would be that I gain the average of the values at position 8, 12, 16 for measured stress_ratio_overload, and measured_stress_ratio_baseline all the other values between 5 and 17. However, when I try to run this code, I get this error:
IndexingError: Too many indexers
How do I use this range to properly index and retrieve the answer I'd like? Any help would be greatly appreciated!
You shouldn't put the mask in the iloc. Since you are using divisor as a standard to find your desired row. You can first add a new column in your dataframe and then slice it.
data['divisor'] = np.arange(len(data)) %4
measured_stress_ratio_overload = data.iloc[range_start:range_finish][data['divisor'] == 0]['test'].mean()
measured_stress_ratio_baseline = data[data['divisor'] != 0].iloc[range_start:range_finish]['test'].mean()
or you can use df.where
measured_stress_ratio_overload = data.iloc[range_start:range_finish].where(data['divisor'] == 0)['test'].mean()
measured_stress_ratio_baseline = data.iloc[range_start:range_finish].where(data['divisor'] != 0)['test'].mean()
I am (re)building up my knowledge of numpy, having used it a little while ago.
I have a question about fancy indexing with multidimenional (in this case 2D) arrays.
Given the following snippet:
>>> a = np.arange(12).reshape(3,4)
>>> a
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
>>> i = np.array( [ [0,1], # indices for the first dim of a
... [1,2] ] )
>>> j = np.array( [ [2,1], # indices for the second dim
... [3,3] ] )
>>>
>>> a[i,j] # i and j must have equal shape
array([[ 2, 5],
[ 7, 11]])
Could someone explain in simple English, the logic being applied to give the results produced. Ideally, the explanation would be applicable for 3D and higher rank arrays being used to index an array.
Conceptually (in terms of restrictions placed on "rows" and "columns"), what does it mean to index using a 2D array?
Conceptually (in terms of restrictions placed on "rows" and "columns"), what does it mean to index using a 2D array?
It means you are constructing a 2d array R, such that R=A[B, C]. This means that the value for rij=abijcij.
So it means that the item located at R[0,0] is the item in A with as row index B[0,0] and as column index C[0,0]. The item R[0,1] is the item in A with row index B[0,1] and as column index C[0,1], etc.
So in this specific case:
>>> b = a[i,j]
>>> b
array([[ 2, 5],
[ 7, 11]])
b[0,0] = 2 since i[0,0] = 0, and j[0,0] = 2, and thus a[0,2] = 2. b[0,1] = 5 since i[0,0] = 1, and j[0,0] = 1, and thus a[1,1] = 5. b[1,0] = 7 since i[0,0] = 1, and j[0,0] = 3, and thus a[1,3] = 7. b[1,1] = 11 since i[0,0] = 2, and j[0,0] = 3, and thus a[2,3] = 11.
So you can say that i will determine the "row indices", and j will determine the "column indices". Of course this concept holds in more dimensions as well: the first "indexer" thus determines the indices in the first index, the second "indexer" the indices in the second index, and so on.
I came across the following Kotlin code:
single(name = walletOkHttpTag) {
createOkHttpClient {
addHeaders(
*mutableListOf<Pair<String, String>>().apply {
add(HeaderKey.ACCEPT to APPLICATION_JSON_HEADER)
if (isDebug || isBeta) {
add(HeaderKey.AUTHORIZATION to BASIC_AUTH_WALLET_STAGE_HEADER)
}
}.toTypedArray()
)
}
}
What does the asterisk * mean that is in front of mutableListOf?
This is the spread operator and it is required to pass an existing array to a vararg function.
When we call a vararg-function, we can pass arguments one-by-one, e.g. asList(1, 2, 3), or, if we already have an array and want to pass its contents to the function, we use the spread operator (prefix the array with *):
Simplified example from the documentation:
val a = arrayOf(1, 2, 3)
val list = listOf(-1, 0, *a, 4)
println(list)
Output:
[-1, 0, 1, 2, 3, 4]
Without the spread operator, the array itself would be added as a single element, resulting in a List<Serializable> with 4 elements:
[-1, 0, [Ljava.lang.Integer;#31befd9f, 4]
What's the best way to do the following in Ramda:
_.range(0, 3, 0);
// => [0, 0, 0]
Thank you.
If you need to repeat the same number n times, then Ori Drori already provided a good answer with repeat.
However if you need to support step, you would have to build a function yourself. (Ramda has a range function but it does not support step.)
So where Lodash would return:
_.range(1, 10, 2);
//=> [1, 3, 5, 7, 9]
You can achieve a similar functionality with Ramda unfold function:
const rangeStep = curry((start, end, step) =>
unfold(n => n < end ? [n, n + step] : false, start));
rangeStep(1, 10, 2);
//=> [1, 3, 5, 7, 9]
You can use R.repeat to create an array of multiple instances of a single item:
const result = R.repeat(0, 3)
console.log(result)
<script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.26.1/ramda.js"></script>
I am trying to change the elements of a C-style array. Using an NSArray/NSMutableArray is not an option for me.
My code is as so:
int losingPositionsX[] = {0, 4, 8...};
but when I enter this code
losingPositionsX = {8, 16, 24};
to change the arrays elements of he array it has an error of: "expected expression" How can I make the copy?
In C (and by extension, in Objective C) you cannot assign C arrays to each other like that. You copy C arrays with memcpy, like this:
int losingPositionsX[] = {0, 4, 8};
memcpy(losingPositionsX, (int[3]){8, 16, 24}, sizeof(losingPositionsX));
Important: this solution requires that the sizes of the two arrays be equal.
You have to use something like memcpy() or a loop.
#define ARRAY_SIZE 3
const int VALUES[ARRAY_SIZE] = {8, 16, 24};
for (int i = 0; i < ARRAY_SIZE; i++)
losingPositionsX[i] = VALUES[i];
Alternatively, with memcpy(),
// Assuming VALUES has same type and size as losingPositions
memcpy(losingPositionsX, VALUES, sizeof(VALUES));
// Same thing
memcpy(losingPositionsX, VALUES, sizeof(losingPositionsX));
// Same thing (but don't use this one)
memcpy(losingPositionsX, VALUES, sizeof(int) * 3);
Since you are on OS X, which supports C99, you can use compound literals:
memcpy(losingPositionsX, (int[3]){8, 16, 24}, sizeof(losingPositionsX));
The loop is the safest, and will probably be optimized into the same machine code as memcpy() by the compiler. It's relatively easy to make typos with memcpy().
I do not know, whether it is a help for you in relation to memory management. But you can do
int * losingPositionsX = (int[]){ 0, 4, 8 };
losingPositionsX = (int[]){ 8, 16, 32 };