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Interaction design: what is the best way to display multiple data points in a single row in a list view?

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Hears!
I am designing a list that needs to accommodate multiple data points in a single row under a single category, and I wonder if there are UI patterns that are more suitable for this need.

I have seen lists that have rows with a large portion of text (such as a paragraph), but it still represents a discrete portion of data, while what I am designing will represent many separate data points (eg Protocol 1, Protocol 2, etc.). )

The user basically needs to quickly and efficiently identify which protocols are associated with their corresponding rules. I've thought about an expanding row interaction, as well as a modal, but both seem a bit click-intensive.

Any ideas? Thanks for taking a look, eager to hear your comments!

Minimum value of a list within another Python list

Hello

I have a situation that I managed to solve by doing loops. I wonder if there is any way to simplify what I did using lambdas or another language function.

I have a list and within it another list, as below:

import numpy as np

lista=()
lista.append((0.21, np.array((0.5, 0.5, 0.667)), 2, 0))
lista.append((0.98, np.array((0.7, 0.9, 0.552)), 1, 1))
lista.append((0.05, np.array((0.5, 0.5, 0.552)), 1, 3))
lista.append((1.18, np.array((0.7, 0.9, 0.552)), 1, 0))

I need to store in another vector the list that contains the smallest value of the first position, which in this case would be the third list (0.05, np.array ((0.5, 0.5, 0.552)), 1, 3).
I did the following:

mins = lista(0)(0)
tmp= lista(0)
vetorfinal =()
for j in range(len(lista)):
    if lista(j)(0) < mins:
        tmp= lista(j)
        mins = lista(j)(0)
vetorfinal.append(tmp)

I was able to solve in an unusual way. I wonder if there is any way to do it with less line of codes.

Thank you

Is it possible for an Infopath form to generate multiple copies of itself uniquely assigned and published in the exchange point list / library?

We are currently using a sharepoint page along with an Infopath form. We create a new form using the + button and complete it and send a separate form for each division to which it belongs (same form, same information, etc., but with a different division name)

Is it possible for the form to generate an identical form for each division (possibly selected from a list) and to be its own unique form instead of having to fill each form individually with the same information over and over again?

We have been able to assign multiple divisions to a form, but we have not had the form create a unique form for each division so that they can sign and complete them.

Method that adds the items of the list of positions and that their sum does not exceed the counter. Java

I need to resolve that this method adds elements to a list and that the elements I add, their sum does not exceed the counter.
For example when I call this method in the main with list = (3,3,5,6) it adds (list, 3), it returns (3) (correct) but when I call in the main this method with list = ( 3,3,5,6) add (list, 4) return me (3,3) and should return me (3). How can i fix it? My code is as follows

 public static PositionList añade (PositionList lista, int contador){
            PositionList resultado = new NodePositionList();
            if(contador<=0) {
                throw new IllegalArgumentException();
            }
            if(list == null || list.isEmpty()) {
                return resultado;
            }
            Position cursor = lista.first();
            Integer suma = 0;
            while(cursor != null && suma < n) {
                if(cursor.element() <= n) {
                    res.addLast(cursor.element());
                    suma = suma + cursor.element();
                }
                cursor = lista.next(cursor);
                }
            return resultado;
            }

python 3.x – Counting frequencies in a list

I have the following code that works to count frequencies in a list. However, the output of this code is NoneType, so I cannot use methods such as .sort, for example, to organize the output to be displayed in descending order of frequency.

Could someone help me solve? I would like the result to be presented as a table or dictionary, but not NoneType.

def frequency(my_list): 

# Creating an empty dictionary  
freq = {} 
for item in my_list: 
    if (item in freq): 
        freq(item) += 1
    else: 
        freq(item) = 1

for key, value in freq.items(): 
    print (" % d : % d"%(key, value))

my_list = (1, 1, 1, 5, 5, 3, 1, 3, 3, 1, 4, 4, 4, 2, 2, 2, 2)

end_list = frequency (my_list)

google sheets: create a list of cell addresses contained in a range

I am looking for a single matrix formula to convert a specific range, such as B2:D4, in a list of addresses (no values!) contained in that range, in this case B2,B3,B4,C2,C3,C4,D2,D3,D4.

Attempts so far

Google did not get useful results.

I tried to count through the matrix and I can extract two lists of numbers corresponding to the row and column numbers, but I could not find a combined formula.

Calcium column:

=ArrayFormula(
   ROW(INDIRECT(SUBSTITUTE(ADDRESS(ROW(),ROW())&":"&ADDRESS(ROW()+
      COLUMNS(B2:D4)
   -1,ROW()),"$","")))-ROW()+
      CELL("col", B2:D4)
)

Calc Row:

=ArrayFormula(
   ROW(INDIRECT(SUBSTITUTE(ADDRESS(ROW(),ROW())&":"&ADDRESS(ROW()+
      ROWS(B2:D4)
   -1,ROW()),"$","")))-ROW()+
      CELL("row", B2:D4)
)

Note that in the formulas above, the ROWS () / COLUMNS () function is used to obtain the size of the range, and CELL () is used to begin the range.

css – How to "delete" WordPress UL categories and make the list repeat on a horizontal line?

When you list categories in WordPress, they are linked through an array in a standard way like this:

             
               catgory1

In fact, the class in question is

apples, pears, bananas, peaches, etc.

Thank you!

python – Understanding the performance optimization list

Loops over large matrices are not really a good idea in Python. That is why his understanding of the original list is not terribly fast.

Its numpy version is loop free, but to my knowledge, np.repeat It actually makes copies of your data, which again is really inefficient. An alternative would be to use np.tile, you may not need to copy the data. But we really don't need to bother, since numpy has a great feature called broadcasting, which often does np.repeat/ /np.tile completely unnecessary Diffusion basically makes np.repeat/tile automatically.

To evaluate performance, I created a more abstract version of the understanding of your list:

def get_valid_op(arr, lowers, uppers):
    return np.asarray((any((val >= lowers) & (val < uppers)) for val in arr))

and also a streaming version

def get_valid_arr(arr, lowers, uppers):
    valid = np.logical_and(arr.reshape(1, -1) >= lowers.reshape(-1, 1), arr.reshape(1, -1) < uppers.reshape(-1, 1))
    return valid.any(axis=0)

The second is practically the same algorithm as its repetition / remodeling code.

With some test data modeled according to its previous description

arr = np.linspace(0, 1000, 70000)
starts = np.linspace(0, 150, 151) * 400
ends = starts + np.random.randint(0, 200, region_starts.shape)  # I assumed non-overlapping regions here

we can first assert all(get_valid_op(arr, starts, ends) == get_valid_arr(arr, starts, ends)) and then time:

%timeit -n 10 get_valid_op(arr, starts, ends)
511 ms ± 5.42 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

%timeit -n 10 get_valid_arr(arr, starts, ends)
37.8 ms ± 3.1 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

An order of magnitude faster. Not bad to start 😉

From working with large matrices (valid has a way of (150, 70000) before the reduction) also has a cost, then I stepped back and went back to the list of loops (just a little).

def get_valid_loop(arr, lowers, uppers):
    valid = np.zeros(arr.shape, dtype=bool)
    for start, end in zip(lowers, uppers):
        valid = np.logical_or(valid, np.logical_and(start <= arr, arr < end))
    return valid

Unlike your understanding of the list, this version now only iterates over the shorter region boundary vectors, which means approximately two orders of magnitude less iterations.

We can then again assert all(get_valid_op(arr, starts, ends) == get_valid_loop(arr, starts, ends)) and time it:

%timeit -n 10 get_valid_loop(arr, starts, ends)
18.1 ms ± 865 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

As the results show, this version is even faster in my "synthetic" reference entries.

In the end, you will have to check the versions in your application and see which one works best.