Is it just me, or is there no binary search function in Phobos? I have a pre-sorted array that I want to search with my own comparator function, but I can't find anything in std.algorithms or std.containers.
Thanks!
Use SortedRange from std.range:
Cribbed from http://www.digitalmars.com/d/2.0/phobos/std_range.html#SortedRange:
auto a = [ 1, 2, 3, 42, 52, 64 ];
auto r = assumeSorted(a);
assert(r.canFind(3));
assert(!r.canFind(32));
Related
I have a for loop but where i has changes by 2 and i want to save a value in a numpy array in each iteration that that changes by 1.
n = 8 #steps
# random sequence
rand_seq = np.zeros(n-1)
for i in range(0, (n-1)*2, 2):
curr_state= i+3
I want to get curr_state outside the loop in the rand_seq array (seven values).
can you help me with that?
thanks a lot
A much simpler version (if I understand the question correctly) would be:
np.arange(3, 15+1, 2)
where 3 = start, 15 = stop, 2 = step size.
In general, when using numpy try to avoid adding elements in a for loop as this is inefficient. I would suggest checking out the documentation of np.arange(), np.array() and np.zeros() as in my experience, these will solve 90% of array - creation issues.
A straight forward list iteration:
In [313]: alist = []
...: for i in range(0,(8-1)*2,2):
...: alist.append(i+3)
...:
In [314]: alist
Out[314]: [3, 5, 7, 9, 11, 13, 15]
or cast as a list comprehension:
In [315]: [i+3 for i in range(0,(8-1)*2,2)]
Out[315]: [3, 5, 7, 9, 11, 13, 15]
Or if you make an array with the same range parameters:
In [316]: arr = np.arange(0,(8-1)*2,2)
In [317]: arr
Out[317]: array([ 0, 2, 4, 6, 8, 10, 12])
you can add the 3 with one simple expression:
In [318]: arr + 3
Out[318]: array([ 3, 5, 7, 9, 11, 13, 15])
With lists, iteration and comprehensions are great. With numpy you should try to make an array, such as with arange, and modify that with whole-array methods (not with iterations).
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>
relatively new to pandas, I have a json and python files:
{"dataset":{
"id": 123,
"data": [["2015-10-16",1,2,3,4,5,6],
["2015-10-15",7,8,9,10,11,12],
["2015-10-14",13,14,15,16,17]]
}}
&
import pandas
x = pandas.read_json('sample.json')
y = x.dataset.data
print x.dataset
Printing x.dataset and y works fine, but when I go to access a sub-element y, it returns a 'buffer' type. What's going on? How can I access the data inside the array? Attempting y[0][1] it returns out of bounds error, and iterating through returns a strange series of 'nul' characters and yet, it appears to be able to return the first portion of the data after printing x.dataset...
The data attribute of a pandas Series points to the memory buffer of all the data contained in that series:
>>> df = pandas.read_json('sample.json')
>>> type(df.dataset)
pandas.core.series.Series
>>> type(df.dataset.data)
memoryview
If you have a column/row named "data", you have to access it by it's string name, e.g.:
>>> type(df.dataset['data'])
list
Because of surprises like this, it's usually considered best practice to access columns through indexing rather than through attribute access. If you do this, you will get your desired result:
>>> df['dataset']['data']
[['2015-10-16', 1, 2, 3, 4, 5, 6],
['2015-10-15', 7, 8, 9, 10, 11, 12],
['2015-10-14', 13, 14, 15, 16, 17]]
>>> arr = df['dataset']['data']
>>> arr[0][0]
'2015-10-16'
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 };
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)