yume | technological platform for developing web | Graphics library

You can use Object.values() to return an array where each object is an item in the array. After that, find your item with Array.find():

Finally, I gave up to fight and decide to refactor API. future_to_promise request static lifetime &str and pulldown_cmark only accept &'a str is a critical problem.

ref: https://qiitadon.com/web/statuses/102710559203790261(written in Japanese)

To refactor API, I made a bold assumption that the parser will produce the same result for the same input. So, I can decide not to hold parser.

As a result, I make such API

Aside from the typo fix above, since its dynamically loaded for lazy-loading purposes, you need to target it in another way.

If you carefully examine:

You can do it with brute force, by converting your rules object into a data.frame, and iteratively comparing LHS/RHS transaction vectors. Here is an example using the grocery.csv dataset:

You are calling engJapDictionary.put(parts[0], parts2[0]); for the first word of the first file with all the words of the second file. Then, you don't put any other words of the first file in the Map, since you finished iterating over the second file in the first run of the inner loop, so line2 = bufferedReader1.readLine() returns null.

You don't need a nested loop, just a single loop that reads a line from both files in each iteration:

I would suggest somehow calculating the normal vectors for each plane you are rendering. Once you have these two vectors - let's say n1 and n2 - it is easy to calculate the angle between the planes with the dot product.

If you aren't familiar with the dot product, dot(n1,n2) where n1 = (x1,y1,z1) and n2 = (x2,y2,z2) would be equal to x1*x2 + y1*y2 + z1*z2. There is another simple identity that says dot(n1,n2) = |v1||v2|cos(a) where || indicates the magnitude of a vector - i.e. |v| = sqrt(x*x + y*y + z*z) if v = (x,y,z) - and a is the angle between the normals which is the angle between the planes. Here is a link to a Mathematics Stack Exchange answer.

In short a = arccos(dot(n1,n2) / |n1||n2|).

If you are interested in learning more about how planes are defined and what the normal vector represents try looking at this.

If you know n1 and n2 are unit vectors then the equation simplifies further to a = arccos(dot(n1,n2)).

So, Matrix is really powerful to solve rotation problems. I flow two steps to set the cube2 rotation and position properly.

• Rotation

  The relationship between cube1,cube2,v1,v2 is: What we want is v2 world rotation equals to v1 opposite world rotation. When two objects have same rotation, they have same rotation matrix(world matrix). We have these formulae:

Try putting the content of the bubble inside a div, and put overflow: scroll on that child div, rather than on the entire bubble.