i have the following document:
{
_id: 12,
item: 'envelope',
qty: ISODate("2021-12-05T00:00:00.000Z"),
arrayField: [ 128, 190, 1 ]
}
and i try to update it using this command
products> db.products.update({_id:12},{$max : { arrayField : [1,190,1879]} })
the output is as follows:
{
acknowledged: true,
insertedId: null,
matchedCount: 1,
modifiedCount: 0,
upsertedCount: 0
}
I don't really understand how the comparison between the existing arrayField and the new one is being done. They are both Arrays, so there should be some kind of comparison on every element, but how exactly does it work?
From the documentation i read this:
With arrays, a less-than comparison or an ascending sort compares the smallest element of arrays, and a greater-than comparison or a descending sort compares the largest element of the arrays. As such, when comparing a field whose value is a single-element array (e.g. 1 ) with non-array fields (e.g. 2), the comparison is between 1 and 2. A comparison of an empty array (e.g. [ ]) treats the empty array as less than null or a missing field.
But i still don't understand exactly... Could someone provide an example in my case?
Thanks in advance
MongoDB takes the min or max, to represent the array in comparisons, for the query operators.
Numbers
{"ar" : [1,2,3]}
(<= ar 1) => (<= min(1,2,3) 3) => (<= 1 3) true
(>= ar 3) => (>= max(1,2,3) 3) => (<= 3 3) true
(= ar 2) => true because it contains the element
For empty arrays either < or > its always false compared to a number
Arrays
(again take the min if < , or max it >)
{"ar" : [1,2,3]}
(<= ar [0 1 2 3]) false because its like min(ar)=1 <min(0,1,2,3)=0
(= ar [1]) false we need all elements =
For the update $max operator
if both are arrays => elements are compared one by one.
max [1 2 3] [5] => [5]
max [1 2] [1 2 -100] => [1 2 -100]
Those are only for the $gte,$gt,$lt,$lte, $eq query operators.
The aggregate ones with same names are strict and don't work like this.
*Its not like complete because we have many types,i think the above are ok, answer might help, it was big to fit in comments.
Related
I have an array with 2 columns. I want to compare the values of the first column, like this:
a[i+1]-a[i]<0.0025.
If this is true I need to delete the row with a[i+1].
This is my first attempt, but it doesnt work.
a = np.delete(a, np.diff(a[:,0])<0.0025, 0)
I get the following error:
ValueError: boolean array argument obj to delete must be one dimensional and match the axis length of 8628
8628 is the length of the array.
Another code i´ve tried is:
a = a[~(np.diff(a[:,0]))<0.0025]
But then I get this error:
TypeError: ufunc 'invert' not supported for the input types, and the inputs could not be safely coerced to any supported types according to the casting rule ''safe''
Can somebody help me with this?
You're on the right track. np.diff(a[:, 0]) < 0.0025 creates an array of length 1 less than 8628, which means when you use it in np.delete that the dimension no longer matches with the original array.
I would go with your second attempt. Using < 0.0025 results in a mask of items that you want to delete, which you need to invert using ~ to get a mask of results you would like to keep. You have to make sure to place your parentheses correctly: ~( np.diff(a[:, 0]) < 0.0025 ). Instead, you can also use >= 0.0025 to make a mask of items you would like to keep.
Lastly, you have to make sure to match the dimensions (given that np.diff results in one less element. You can do this by prepending True to signify that you always want to keep the first value. One way to do that is using np.r_.
Final code:
import numpy as np
a = np.random.rand(8628, 2) # Your array here
result = a[ np.r_[True, np.diff(a[:, 0]) >= 0.0025] ]
Detailed example:
Consider the array: [ 1, 3, 2, 5, 3]
np.diff creates: [ 2, -1, 3, -2]
Using threshold creates: [True, False, True, False]
Note that when the next element in the original array is less than the previous,
the thresholding result in False too.
Finally, because there are now 4 values instead of 5, we prepend True.
This has the effect of always including the first element in the result:
Original: [ 1, 3, 2, 5, 3]
Mask [True, True, False, True, False]
^^^^ ~~~~ ~~~~~ ~~~~ ~~~~~
Then using boolean indexing, we get the elements where
the mask contains True to obtain the final result:
[1, 3, 5]
Working on a project to recreate a game Mastermind. I need to compare two arrays, and running into some struggles.
I need to output two integers for the flow of the game to work,
the first integer is the number of correct choices where the index matches. The code I have for this appears to be working
pairs = #code.zip(guess)
correct_position_count = pairs.select { |pair| pair[0] == pair[1] }.count
Where pairs is equal to a 4 element array and the guess is also a 4 element array
The second part I am having a bit of trouble with on how to do the comparison and return an array. The integer should represent where the two arrays index don't match (the above code block but !=) and confirm whether the guess array excluding any exact index matches has any elements included with the code array once again excluding the exact index matches.
Any help would be greatly appreciated!
I am not completely sure to understand your problem but if I understood well, you've two arrays, solution with the solution and guess with the current guess of the player.
Now, let's assume that the solution is 1234 and that the guess is 3335.
solution = [1, 2, 3, 4]
guess = [3, 3, 3, 5]
an element by element comparison produces an array of booleans.
diff = guess.map.with_index { |x,i| x == solution[i] }
# = [false, false, true, false]
Now, you can easily compute the number of good digits diff.count true and the number of wrong digits diff.count false. And, in case you need the index of the false and/or true values you can do
diff.each_index.select { |i| diff[i] } # indexes with true
# = [2]
diff.each_index.select { |i| !diff[i] } # indexes with false
# = [0, 1, 3]
You can count all digit matches ignoring their positions and then subtract exact matches.
pairs = #code.zip(guess)
correct_position_count = pairs.select { |pair| pair[0] == pair[1]}.count
any_position_count = 0
code_digits = #code.clone # protect #code from modifying
guess.each do |digit|
if code_digits.include?(digit)
code_digits.delete_at(code_digits.find_index(digit)) # delete the found digit not to count it more than once
any_position_count += 1
end
end
inexact_position_count = any_position_count - correct_position_count
puts "The first value: #{correct_position_count}"
puts "The second value: #{inexact_position_count}"
I want to get a range from my sheet. As recommended in Best practices, I am trying to get a array and manipulate it, but I'm confused:
const ss = Spreadsheet.getActive(),
sh = ss.getSheetByName("Sheet1"),
rg = sh.getRange("A1:C1"),//has 1,2,3
values = rg.getValues();
console.log(values);
The logs show
[[1,2,3]]
As you can see I got all three elements. But, when I log the length of the array(array.length), it is just 1(instead of 3). When I test existence of a element using .indexOf or .includes, It says -1 or false.
const values = /*same as logged above*/[[1,2,3]];
console.log(values.indexOf(2));//got -1 expected 1
console.log(values.includes(1));//got false expected true
Why?
I have the same issue with setValues().
rg.setValues([1,2,3]);//throws error
The error is
"The parameters (number[]) don't match the method signature for SpreadsheetApp.Range.setValues."
My specific Question is: What exactly does getValues() return? Is it a special kind of array?
Documentation excerpts:
From The official documentation, getValues() returns
a two-dimensional array of values,
It ALWAYS returns a two dimensional array of values.
One dimensional array is
[1,2,3]
Two dimensional array is
[[1,2,3]]
//or
[[1], [2], [3]]
There is/are array(s) inside a array.
indexed by row, then by column.
It is indexed by row first: i.e., The outer array has rows as inner array. Then each inner array has column elements. Consider the following simple spreadsheet:
A
B
C
1>
1
2
3
2>
2
3
4
3>
3
4
5
A1:A3 contains 3 rows and each row contains 1 column element. This is represented as [[1],[2],[3]]. Similarly, The following ranges represent the following arrays. Try to guess the array structure based on the A1 notation:
A1Notation
Number of Rows
Number of columns
Array Structure
array.length
array[0].length
A1:A3
3
1
[[1],[2],[3]]
3
1
A1:C1
1
3
[[1,2,3]]
1
3
A1:B2
2
2
[[1,2],[2,3]]
2
2
B1:C3
3
2
[[2,3],[3,4],[4,5]]
3
2
A2:C3
2
3
[[2,3,4],[3,4,5]]
2
3
Note how the two dimension provides direction.
See live visualization below:
/*<ignore>*/console.config({maximize:true,timeStamps:false,autoScroll:false});/*</ignore>*/
const test = {
'A1:A3': [[1], [2], [3]],
'A1:C1': [[1, 2, 3]],
'A1:B2': [
[1, 2],
[2, 3],
],
'B1:C3': [
[2, 3],
[3, 4],
[4, 5],
],
'A2:C3': [
[2, 3, 4],
[3, 4, 5],
],
};
Object.entries(test).forEach(([key, value]) => {
console.log(`The range is ${key}`);
console.table(value);
console.info(`The above table's JavaScript array notation is ${JSON.stringify(value)}`)
console.log(`=================================`);
});
<!-- https://meta.stackoverflow.com/a/375985/ --> <script src="https://gh-canon.github.io/stack-snippet-console/console.min.js"></script>
The values may be of type Number, Boolean, Date, or String, depending on the value of the cell.
In the above example, We have Spreadsheet Number type elements converted to JavaScript number type. You can check spreadsheet type using =TYPE(). Corresponding JavaScript type reference is here
Empty cells are represented by an empty string in the array.
Check using
console.log(values[0][0]==="")//logs true if A1 is empty
Remember that while a range index starts at 1, 1, the JavaScript array is indexed from [0][0].
Given the two dimensional array structure, to access a value, two indexes of format array[row][column] is needed. In the above table, if A2:C3 is retrieved, To access C3, Use values[1][2]. [1] is second row in range A2:C3. Note that the range itself starts on second row. So, second row in the given range is row3 [2]is third column C.
Notes:
Warning:
Retrieved values from a range is always two dimensional regardless of the range height or width(even if it is just 1). getRange("A1").getValues() will represent [[1]]
setValues() will accept the same array structure corresponding to the range to set. If a 1D array is attempted, the error
The parameters (number[]/string[]) don't match the method signature for SpreadsheetApp.Range.setValues.
is thrown.
If the array does NOT exactly correspond to the range being set,i.e.,if each of the the inner array's length does not correspond to the number of columns in the range or the outer array's length does not correspond to the number of rows in the range being set, The error similar to the following is thrown:
The number of columns in the data does not match the number of columns in the range. The data has 5 but the range has 6.
Related answers to the above error:
https://stackoverflow.com/a/63770270
Related Search
indexOf/includes uses strict type checking. They won't work when you compare primitives against array objects. You can use Array.flat to flatten the 2D array to a 1D one. Alternatively, Use a plain old for-loop to check something.
const values = [[1,2,3]].flat();//flattened
console.log(values.indexOf(2));//expected 1
console.log(values.includes(1));//expected true
References:
Basic reading
MDN Arrays guide
I have a numpy array
my_array = np.array([[1,2,3,4],[5,6,7,8],[0,0,0,0],[1,2,3,4],[0,0,0,0],[0,0,0,1]])
and I would like to get all index when array contains only zero values :
index 2 -> [0,0,0,0]
index 4 -> [0,0,0,0]
Discussion with the similar problem exists : Find indices of elements equal to zero in a NumPy array
but in this solution we get values equal to zero, instead of get array with zero as I want.
Thank for your help.
You can use np.argwhere with np.all to get indices of rows where all elements == 0:
In [11] np.argwhere((my_array == 0).all(axis=1))
Out[11]:
array([[2],
[4]], dtype=int64)
Or np.nonzero instead of np.argwhere gives slightly nicer output:
In [12] np.nonzero((my_array == 0).all(axis=1))
Out[12]: (array([2, 4], dtype=int64),)
Given two arrays of equal size, how can I find the number of matching elements disregarding the position?
For example:
[0,0,5] and [0,5,5] would return a match of 2 since there is one 0 and one 5 in common;
[1,0,0,3] and [0,0,1,4] would return a match of 3 since there are two matches of 0 and one match of 1;
[1,2,2,3] and [1,2,3,4] would return a match of 3.
I tried a number of ideas, but they all tend to get rather gnarly and convoluted. I'm guessing there is some nice Ruby idiom, or perhaps a regex that would be an elegant answer to this solution.
You can accomplish it with count:
a.count{|e| index = b.index(e) and b.delete_at index }
Demonstration
or with inject:
a.inject(0){|count, e| count + ((index = b.index(e) and b.delete_at index) ? 1 : 0)}
Demonstration
or with select and length (or it's alias – size):
a.select{|e| (index = b.index(e) and b.delete_at index)}.size
Demonstration
Results:
a, b = [0,0,5], [0,5,5] output: => 2;
a, b = [1,2,2,3], [1,2,3,4] output: => 3;
a, b = [1,0,0,3], [0,0,1,4] output => 3.
(arr1 & arr2).map { |i| [arr1.count(i), arr2.count(i)].min }.inject(0, &:+)
Here (arr1 & arr2) return list of uniq values that both arrays contain, arr.count(i) counts the number of items i in the array.
Another use for the mighty (and much needed) Array#difference, which I defined in my answer here. This method is similar to Array#-. The difference between the two methods is illustrated in the following example:
a = [1,2,3,4,3,2,4,2]
b = [2,3,4,4,4]
a - b #=> [1]
a.difference b #=> [1, 3, 2, 2]
For the present application:
def number_matches(a,b)
left_in_b = b
a.reduce(0) do |t,e|
if left_in_b.include?(e)
left_in_b = left_in_b.difference [e]
t+1
else
t
end
end
end
number_matches [0,0,5], [0,5,5] #=> 2
number_matches [1,0,0,3], [0,0,1,4] #=> 3
number_matches [1,0,0,3], [0,0,1,4] #=> 3
Using the multiset gem:
(Multiset.new(a) & Multiset.new(b)).size
Multiset is like Set, but allows duplicate values. & is the "set intersection" operator (return all things that are in both sets).
I don't think this is an ideal answer, because it's a bit complex, but...
def count(arr)
arr.each_with_object(Hash.new(0)) { |e,h| h[e] += 1 }
end
def matches(a1, a2)
m = 0
a1_counts = count(a1)
a2_counts = count(a2)
a1_counts.each do |e, c|
m += [a1_counts, a2_counts].min
end
m
end
Basically, first write a method that creates a hash from an array of the number of times each element appears. Then, use those to sum up the smallest number of times each element appears in both arrays.