I managed to create a data connection in InfoPath with the hope of retrieving the information of the current user, the name to be displayed, etc.

However, whenever you drag the desired field to the InfoPath form, a Repeating Section is created that includes all the entries for that field.

How can I isolate the information so that it is only for the current user?

# Tag: list

## ComboBox mouse scroll list in 64-bit Excel 2016

I have a drop-down list of ComboBox that has more than 150 lines. I was wondering if there is a way to control the scroll with the mouse wheel over the lines so you do not need to use the bar on the right.

## Where to see a list of all non-special MediaWiki web pages?

I get the impression that in MediaWiki there are two main groups of **no article** Y **not speak** websites:

Special web pages (such as `special: categories`

)

Media-Wiki "Media-Wiki" web pages (as `media-wiki: common.css`

).

While I can see a list of all the special web pages in `special: special_pages`

, I miss where I can find a list of `media-wiki: media_wiki_pages`

.

Do you know where you can find in the navigation?

## 30 days to build your largest email list for $ 2

#### 30 days to build your largest email list

**Now you can generate massive sales and profits with your own sensitive email list! Keep reading to discover a simple 30-day plan for a larger email list!**

The number one thing that almost all successful online entrepreneurs have in their favor is that they have an email list. This is a list of subscribers who chose to receive more information from you, get access to free training or people who bought you something.

At one time or another, we have all reached an email list. Email marketing continues to be the number one way to engage and follow up with your audience, from the largest online retailers to the solopreneurs who run a business from your living room.

Social networks have definitely changed the game a bit, but email marketing is here to stay. What has changed is how people access their emails. Therefore, do not let anyone try to trick you into thinking that email marketing is dead.

In fact, some companies, even the most important ones, credit more than 75% of their sales directly to email marketing.

If you have an email list and create a product, have a training program, a service, a physical product or a webinar about which you want to talk to people; Then just send an email to your list to get instant results! Do not wait any longer for your paid ads to be approved or trust affiliates to mail you.

Paid advertising and affiliates are a great way to get traffic, but having your own email list is even better! That's why during the next 30 days, this guide will teach you the same steps that experts used to create an email list that pays you repeatedly!

With 30 days to a larger list, you're on track to earn a lot online and build that asset you need to help keep your income consistent. There has been a constant in the world of marketing, and that's email! You must have an email list if you want to do it!

Instead of dividing this into chapters like a normal book, it will have each section of the process with its own section so you can easily follow it!

**Product terms**

[YES]It can be used for personal use

.

## List manipulation: Find the 3D region that contains the source delimited by given planes

I am writing a code to generate the Wigner-Seitz cell of the reciprocal network for a given set of network translation vectors. For example, consider the cubic centered body network (BCC), whose base translation vectors are given by

```
a1 = {-1, 1, 1} / 2;
a2 = {1, -1, 1} / 2;
a3 = {1, 1, -1} / 2;
```

The reciprocal basis vectors are defined according to

```
d = 2 Pi;
v = a1. (a2 [Cross]a3);
b1 = d / v (a2 [Cross]a3);
b2 = d / v (a3 [Cross]a1);
b3 = d / v (a1 [Cross]a2);
```

The reciprocal network is then defined by the set of reciprocal network vectors, the set of all linear combinations of integer multiples of reciprocal base vectors, that is,

$$ vec {G} = n_1 vec {b} _1 + n_2 vec {b} _2 + n_3 vec {b} _3, qquad n_i in mathbb {Z} $$

The Wigner-Seitz cell (in this case, the first Brillouin zone) is defined as the region that contains the origin that is bounded by the perpendicular bisection planes of the reciprocal network vectors. In general, we can achieve this by considering only the first, second and perhaps the third point of the network closest to the origin. In the case of BCC, for example, the following vectors will suffice:

```
recipvecs =
Select[Flatten[
Table[n1 b1 + n2 b2 + n3 b3, {n1, -1, 1}, {n2, -1, 1}, {n3, -1, 1}], two],
Rule[#] <= 2 d &];
```

**Question:** Given these vectors, how can I build the Wigner-Seitz cell?

For example, one possibility is to construct the equations for all planes.

```
plans = ({x, y, z} - (# / 2)). # == 0 & / @ reciplattice
```

(note that there is a redundancy for the origin, which simply gives `True`

, this can be eliminated). Now, the problem will be to rewrite each of these equations as an inequality in such a way that the half space defined by the inequality contains the origin. I do not think it's too difficult, but not all equations can be solved for any of the coordinates, for example. we can not solve all the equations for $ z $, I like it

```
Solve[#, z] & / @ blueprints
```

Some of the equations must be solved for $ x $ or $ and $ Before becoming inequalities. I think I could find a brute force solution but I hope there is something more elegant.

Ultimately, I would like to obtain the inequalities that define the region in order to visualize it with `RegionPlot3D`

and use it to `Select`

points of a mesh.

**To update:**

So far I made a single `Yes`

build to solve the equations for anyone $ z $, $ x $or $ and $,

```
sols = flat
The
Yes[sz=Solve[sz=Solve[sz=Resolver[sz=Solve[#, z]; sz! = {}, sz,
Yes[sx = Solve[#, x]; sx! = {}, sx, Resolve[#, y]]]& / @ blueprints]
```

which results in the following list of rules

```
{z -> -1, z -> -1 - y, z -> -1 + x, y -> -1, x -> 1 + y, x -> 1, z -> -1 - x, z -> -1 + y, x -> -1 - y, x -> 1 - y, z -> 1 + y, z -> 1 - x, x -> -1, x -> -1 + y, and -> 1, z -> 1 + x, z -> 1 - y, z -> 1}
```

Now I just need to convert each of these into $ LHS <RHS $ Y $ LHS> RHS $ and test which contains the origin, but I still do not know how to achieve it.

## html: the selection panel does not contain numbers from the entry list

Choose the file number:

Where option is the list of numbers (from the flask render_template (& # 39; my.html & # 39 ;, option =[1, 2, 3])).

The variable & # 39; name & # 39; It is necessary to:

```
```

## python – Please convert to the comprehension list

```
def to_one_hot (x):
"" "Converts an integer array to a binary array.
Parameters
----------
x: array_like of int
One-dimensional matrix containing integers from 0 to N.
Returns
-------
out: list of int list
Binary matrix of input representation.
Examples
--------
>>> to_one_hot ([3, 2, 1, 0])
[[0, 0, 0, 1], [0, 0, 1, 0], [0, 1, 0, 0], [1, 0, 0, 0]]>>> to_one_hot ([2, 2, 1, 0])
[[0, 0, 1], [0, 0, 1], [0, 1, 0], [1, 0, 0]]>>> to_one_hot ([3, 3, 1, 0])
[[0, 0, 0, 1], [0, 0, 0, 1], [0, 1, 0, 0], [1, 0, 0, 0]]""
result = [[[0]* (max (x) +1)]for k in range (len (x)))]result = [[0]* (max (x) +1)]* len (x)
for i, j in zip (range (len (x)), x):
result[i][j] = 1
return result
```

## forms: name of a control with a drop-down menu, text boxes and a list

I'm creating a control with a drop-down menu, text boxes and a list; It will look a bit like this:

When you click on the `+`

button, the values of the text boxes will be added to the list. The category of the added values will be the option selected from the drop-down menu. That same category would be removed from the options.

Clicking on the `X`

On the other hand, the button would remove the item from the list and add the category to the drop-down options.

Is there a generic name for this type of collective controls? If there is not, what would be a good name for this?

## list manipulation: merge two rows with rounding of their first elements

```
lst = {{-0.410927, {{11, 16}, {12, 16}, {15, 5}, {15, 6}}},
{-0.41, {{11, 16}, {12, 16}, {15, 5}, {15, 6}}},
{-0.45123, {{11, 16}, {12, 16}, {15, 5}, {15, 6}}},
{-0.410927, {{9, 14}, {10, 14}, {13, 7}, {13, 8}}}};
Group by[Round[First @ #, .001]&]@ lst
```

<|-0.411 -> {{-0.410927, {{11, 16}, {12, 16}, {15, 5}, {15,

6}}}, {-0.410927, {{9, 14}, {10, 14}, {13, 7}, {13,

8}}}},

-0.41 -> {{-0.41, {{11, 16}, {12, 16}, {15, 5}, {15,

6}}}}

-0.451 -> {{-0.45123, {{11, 16}, {12, 16}, {15, 5}, {15,

6}}}} |>

```
Group by[Round[First @ #, .01] &]@ lst
```

<|-0.41 -> {{-0.410927, {{11, 16}, {12, 16}, {15, 5}, {15, 6}}}, {-0.41, {{11, 16}, {12, 16}, { 15, 5}, {15, 6}}}, {-0.410927, {{9, 14}, {10, 14}, {13, 7}, {13, 8}}}},

-0.45 -> {{-0.45123, {{11, 16}, {12, 16}, {15, 5}, {15, 6}}}} |>

## list manipulation: why does a table with constants defined in its index calculate 10X slowest?

The problem lies mainly in the interior `Table`

:

```
Synchronization[Total[Table[Total[Table[data[[i]], {i, j, 10 + j}]], {j, 1, length[data] - 5 * 10}]]]m = 10;
Synchronization[Total[Table[Total[Table[data[[i]], {i, j, m + j}]], {j, 1, length[data] - 5 * 10}]]]
```

{0.366407, 5.50276 * 10 ^ 6}

{8.01738, 5.50276 * 10 ^ 6}

I think the reason is this:

Because the global variable `subway`

theoretically it could change its value during the calculations, the body of the external table can not be compiled (without calls to MainEvaluate). At least, the JIT compiler does not analyze the body of the external loop sufficiently thoroughly to decide that `subway`

it will not change

You can help the JIT compiler by using `With`

:

```
With[{m = 10},
Synchronization[Total[Table[Total[Table[data[[i]], {i, j, m + j}]], {j, 1, length[data] - 5m}]]]]
```

{0.369601, 5.5049 * 10 ^ 6}

By focusing on the title of the publication, I have overlooked the question of how to do it faster. Here is my proposal (`do`

) vs. the one of the OPs (`a`

) and Carl (`second`

):

```
a = with[{m = 10},
Total[
Table[Total[Table[data[[i]], {i, j, m + j}]]{j, 1,
Length[data] - 5m}]]]; // Repeated time // First
b = Total @ ListCorrelate[ConstantArray[1., m + 1],
data[[;; -50 + m - 1]]]; // Repeated time // First
c = Plus[
Range[1., m].data[[1 ;; m]],
(m + 1) Total[data[[m + 1 ;; -5*m - 1]]],
Distance[N@m, 1., -1].data[[-5 m ;; -4 m - 1]]]; // Repeated time // First
a - b
a - c
```

0.28

0.017

0.0018

True

True