redmoddata/lines_train_2M · Datasets at Hugging Face

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Delfina (noun) Del-fina (from-fine)
1. Daughter of the Roman god Apollo and Daphne.
Delfina is the goddess ruler of the sun. In heathen theology; a cherisher and protector of the polite arts, Inherited Like Apollo she was the god of light, medicine, eloquence, music and beauty.
2 Essentially a female name (Latin) Spanish or European upper class
3. Delfina is from the finest, ultrafine, superfine, and emotionally stable. "Fina" fine in every aspect.
4. All that is good, classy, real and pure.
5. The modern day Delfina has an air of authority; it is an innate right that suits her completely.
6. Anything that comes from the very finest, of superior quality, of high regard
7. from the most precious of all things considered
8. A natural beauty illuminates her surroundings; she is from the finest stock.
9. Fine facial appearance, captivating eyes, long nose, great body, round tight behind, small waist.
It was obvious by the lavish clothes they wore that they lived a delfina lifestyle.
by bradcoop August 28, 2010
71 6
Take the 2-minute tour ×
I am writing a script and started working with the install command (for copying files) and it is not working. CMake configure/generate does not show any errors (i.e. it does not stop and no warnings/errors show related to this command) and the command does not seem to be working because I don't see any files being copied.
Since I am new, I am wondering:
• How can I tell that install failed (perhaps the source directory was wrong, or the destination directory was wrong)? It appears to be failing silently
• Are there error codes I can check to see what went wrong?
• Ans lastly when is install called? When I click configure? Or when the project is built?
I am on Windows.
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1 Answer
When you add an install command to your CMakeLists.txt, you get a new target created called "install".
In order to actually install the chosen files/targets, you need to build this install target. It's not automatically built as part of the "ALL" target.
For example, if you're using Visual Studio, the "INSTALL" target should appear in the "CMakePredefinedTargets" folder of the Solution Explorer. Just selecting this target and building it will cause the solution to be built and the selected items installed.
If any part of the build or install process fails, the notifications should then be apparent.
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That makes a lot of sense. Either I am not looking hard enough or the documentation does a poor job at describing these details. – Samaursa Mar 24 '13 at 1:07
This tutorial might help a bit? – Fraser Mar 24 '13 at 2:19
It does! Thanks! – Samaursa Mar 24 '13 at 15:51
add comment
Your Answer
Take the 2-minute tour ×
I have an object $this->user which is of model User. This object is populated by $this->Auth->user in my app controller like so:
$this->user = ClassRegistry::init('User');
Works like a charm. If I print_r out $this->user in my controller it gives me:
User Object ( [validate] => Array ( blah blah blah
A typical object. Now I have a Group model which belongs to a User, and users have many groups. These variables are properly set in the models. Now I want to find all Groups for this particular user who is logged in. So I tried this:
$groups = $this->user->Group->find('list', array('fields'=>array('id', 'group_name')))
The key is that I want to use $this->user to automatically filter the Group query based on the owner_id in $this->user. It makes sense to me that if I've got a specific object representing a user and I do a Group query based on that user ... it should only return the relevant groups.
The problem is that $groups contains all of the entries in the Groups table, rather than obviously the ones I only want from the current user. I don't see why I would need to add a "conditions"=>"user_id"=$this->Auth->user('id') parameter to the find function because I've already specified what user I'm using via the model chain.
Any ideas why this is not working? The sql statment it runs is simply a SELECT on Groups WHERE 1 = 1 (so not filtering at all).
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1 Answer
up vote 0 down vote accepted
No, the object User really acts more like a class than a object. You can say Cake doesn't fully implement Active Record pattern (I think Cake 3.0 may fix that, not sure). So yes, you still need to set the condition for the find. And you don't have to set($this->Auth->user);
You are probably not very familiar with Cake: in Cake, you hardly ever have to instantiate Model objects at all. They are created for you based on the current controller and model relationships that you specify.
share\|improve this answer
OK thanks for your answer. I am still learning cake, yes. The only reason I had that variable instantiated is because I wanted to use it in this manner - as that is not possible then you're right - I have no use for it. – MikeMurko Aug 26 '11 at 19:21
add comment
Your Answer
Subscribe English
look up any word, like fuck:
Scapulation dumping refers to the process of dumping scapula’s. Little is known about Scapula’s (from the latin word Scapulaiincredabilous) except that they are found at incredible depths within the worlds oceans. It is within their home environment that they creep and crawl over the ocean floors gathering the bubbles from shark farts.
Deep sea trawlers net the scapulas, extract the shark farts and encapsulate them in a vat of spirits which are ultimately sold to the manufacturers of spirit levels and used for the bubbles found within those levels.
Extreme caution must be applied throughout the action to avoid the shark farts entering the seamen’s blood stream and causing “the bends”.
The Scapula’s are not harmed during the process and are dumped back into the sea once their valuable cargo’s are extracted hence the expression “scapulation dumping”.
Prof J.S. Miller
Scapulation Dumping refers to dumping scapula's and is associated with production of Pioneer Brickies Levels
by Ormbo January 22, 2008
2 3
Words related to Scapulation Dumping refers to dumping scapula's:
dumping farts levels scapula sharks
Take the 2-minute tour ×
This seems like an ideal place to ask a question that has been keeping me confused for a little while now. But I apologize if I've posted in the wrong area.
My question is that I don't seem to understand the temperature trend that I am seeing from my temperature sensors. I'll start of with my setup: Essentially, I have a metal box with two temperature sensors. 1) Is mounted right at the base of the box [Temp Sensor #1] 2) The second temperature sensor is mounted 2 inches from the top of the box [Temp Sensor #2]. The box is about 6 inches in height.
I've placed a heating pad at the base of the box (Temp Sensor #1 lies right in the center of this pad. A graphic showing my setup:
I've also implemented a simple on/off temperature controller, that senses when the temperature goes above a certain set-point and turns off (hence you see the highs/lows for Temp Sensor #1).
As you can see from the image, the lower sensor (blue) has the peaks/troughs corresponding to when the heater turns on/off. The heater gets triggered every time the sensing temperature goes below the set point. What I don't understand is why the top sensor (red) has a periodically decreasing trend (it doesn't have highs/lows similar to the bottom sensor)? It doesn't seem to be affected by the heater turning on at all? Even though it is merely 4" away from the heater inside the metal enclosure.
I understand that Sensor#1 is probably changing immediately due to the heater very quickly affecting the metal base temperature via conduction. Whereas the second sensor is probably measuring the air around the metal enclosure at the top, and since air is an insulator, it takes longer to heat up. But there should be at-least some highs and lows I'd imagine. The continuous decreasing trend doesn't make any sense ...
Then, I suspected that perhaps my second temperature sensor was damaged. But that wasn't the case. I've tested both sensors and they work fine. Also here is a graph of the temperature trend, when I place the enclosure (with the sensors) in the freezer with no heater action. Intuitively as you can imagine, there is merely a decreasing trend for both sensors (shown below) due to the effect of the freezer:
Any suggestions please as to why I notice no temperature variation at Sensor #2 location when the heater turns on/off?
share\|improve this question
Here is the trend with no heater and the freezer merely cooling: i47.tinypic.com/2430vp2.png , I couldn't post it in the original post since I do not have sufficient rep points. The heater pad type, I am using is this: winemakersdepot.com/Brewers-and-Wine-Making-Heat-Pad-P700.aspx – c0d3rz Jan 31 '13 at 0:32
I see when the heater is turned on an off, but when is the cooling activated? – mdma Jan 31 '13 at 1:32
What temp are you trying to stabilize at and is this metal enclosure insulated? Especially the lid. – brewchez Feb 2 '13 at 22:19
You've also inappropriately applied a linear fit to data that isn't behaving that way. You state that its a continuous decreasing trend. But it isn't, it has stabilize half way through the data collection period. – brewchez Feb 2 '13 at 22:24
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1 Answer
up vote 1 down vote accepted
It's to do with thermal inertia. If you look closely at the graph you'll see there are highs and lows for sensor #2 also - just much smaller than sensor #1, and they have the same period (time interval) as sensor #1, indicating they stem from the same heating oscillation.
You're of course right when you say that air is a poor conductor, and so it will essentially dampen the effect of the heater - there are rises and falls, just over a smaller, dampened range.
I'm not entirely clear on when the freezer switches on and why it would be on when the heater is also on, but it seems that's the case from the graph. Another reason for the imbalance is that the freezer has a far greater cooling effect than the 25W heater can heat. Most freezers are in the order of 120W or more, and have a Coefficient of Performance (CoP) of 2 or more when chilling to beer temperature, so you're getting about 240W of real cooling power vs 25W of heating power (resistance heating has a CoP <= 1) - so a 10:1 difference.
I imagine the main reason you're seeing any fluctuation in sensor 2 at all is because of the box, which confines the convection of heated air to within the box. I imagine if you removed the box and did the same thing, you'd see very little change in sensor #2 because of the large volume of cooler air surrounding it.
share\|improve this answer
Hi mdma: The freezer is always switched on, it is never turned off. Also, I'm using a 120W heater instead, so the heating/cooling difference is not that large I think. If I go with the idea of thermal inertia, why is that the lower sensor (Sensor #1) changes so quickly then? Shouldn't it face the same problem? – c0d3rz Jan 31 '13 at 2:56
The lower sensor changes quickly because the rate of conduction through contact with #1 is much quicker than conduction through the air to #2. – mdma Jan 31 '13 at 2:59
Hmm, just one last question about this CoP. Why is that the CoP is greater than 1 in the case of cooling? – c0d3rz Jan 31 '13 at 20:59
The CoP for compressor/inverter systems like fridges and heating pumps because they are just moving heat from one place to another, so the net heat output can be higher than the energy consumed. With a resistance based heat pad, this turns electrical energy into heat, and so the amount of heat produced can never be more than the electricity consumed. – mdma Jan 31 '13 at 21:09
add comment
Your Answer
• By
Rockstar Games
Image from L.A. Noire
Rockstar Games last week released L.A. Noire, a hypnotic detective thriller developed by Australian studio Team Bondi. The videogame has been called groundbreaking for the facial-recognition technology used to create its characters, and the resulting psychological dimensions of gameplay. In witness questioning and interrogation sequences, the player, as young detective Cole Phelps, tries to ascertain the truths and lies in what he’s told. Because L.A. Noire’s cast of characters feature the recorded facial performances of its actors, rather than hand-animated abstractions of feeling, the gameplay carries the emotional ambiguity and weight of a cop show or film noir. We spoke via email with Jeronimo Barrera, VP of product development at Rockstar, about the process of building the game.
Speakeasy: The team used maps from the Works Progress Administration, archival photographs from newspapers, and aerial photos as reference material for the environment. Can the eight-square-mile section of Los Angeles recreated in the game be considered documentary or used as research material in its own right?
Jeronimo Barrera: We went to a great deal of effort in every area, from the details of the interiors of certain key buildings, licensing classic billboards, and recreating former landmarks, but it’s not a block-for-block recreation—we are trying to make something that feels real, not that replicates reality. We have to make some concessions to game design, alongside more pedestrian things like building licensing. A great example of the lighter side of the issues involved is the fact that in 1947, all the palm trees in L.A. were still roughly head-high. But when they were designed that way in the game, they had the opposite effect to being realistic—they actually distracted players from the experience, so we designed them at full height. We also designed the game to be slightly less smoggy than L.A. was at the time.

Delfina (noun) Del-fina (from-fine)

1. Daughter of the Roman god Apollo and Daphne.

Delfina is the goddess ruler of the sun. In heathen theology; a cherisher and protector of the polite arts, Inherited Like Apollo she was the god of light, medicine, eloquence, music and beauty.

2 Essentially a female name (Latin) Spanish or European upper class

3. Delfina is from the finest, ultrafine, superfine, and emotionally stable. "Fina" fine in every aspect.

4. All that is good, classy, real and pure.

5. The modern day Delfina has an air of authority; it is an innate right that suits her completely.

6. Anything that comes from the very finest, of superior quality, of high regard

7. from the most precious of all things considered

8. A natural beauty illuminates her surroundings; she is from the finest stock.

9. Fine facial appearance, captivating eyes, long nose, great body, round tight behind, small waist.

It was obvious by the lavish clothes they wore that they lived a delfina lifestyle.

by bradcoop August 28, 2010

71 6

Take the 2-minute tour ×

I am writing a script and started working with the install command (for copying files) and it is not working. CMake configure/generate does not show any errors (i.e. it does not stop and no warnings/errors show related to this command) and the command does not seem to be working because I don't see any files being copied.

Since I am new, I am wondering:

• How can I tell that install failed (perhaps the source directory was wrong, or the destination directory was wrong)? It appears to be failing silently

• Are there error codes I can check to see what went wrong?

• Ans lastly when is install called? When I click configure? Or when the project is built?

I am on Windows.

share|improve this question

add comment

1 Answer

When you add an install command to your CMakeLists.txt, you get a new target created called "install".

In order to actually install the chosen files/targets, you need to build this install target. It's not automatically built as part of the "ALL" target.

For example, if you're using Visual Studio, the "INSTALL" target should appear in the "CMakePredefinedTargets" folder of the Solution Explorer. Just selecting this target and building it will cause the solution to be built and the selected items installed.

If any part of the build or install process fails, the notifications should then be apparent.

share|improve this answer

That makes a lot of sense. Either I am not looking hard enough or the documentation does a poor job at describing these details. – Samaursa Mar 24 '13 at 1:07

This tutorial might help a bit? – Fraser Mar 24 '13 at 2:19

It does! Thanks! – Samaursa Mar 24 '13 at 15:51

add comment

Your Answer

Take the 2-minute tour ×

I have an object $this->user which is of model User. This object is populated by $this->Auth->user in my app controller like so:

$this->user = ClassRegistry::init('User');

Works like a charm. If I print_r out $this->user in my controller it gives me:

User Object ( [validate] => Array ( blah blah blah

A typical object. Now I have a Group model which belongs to a User, and users have many groups. These variables are properly set in the models. Now I want to find all Groups for this particular user who is logged in. So I tried this:

$groups = $this->user->Group->find('list', array('fields'=>array('id', 'group_name')))

The key is that I want to use $this->user to automatically filter the Group query based on the owner_id in $this->user. It makes sense to me that if I've got a specific object representing a user and I do a Group query based on that user ... it should only return the relevant groups.

The problem is that $groups contains all of the entries in the Groups table, rather than obviously the ones I only want from the current user. I don't see why I would need to add a "conditions"=>"user_id"=$this->Auth->user('id') parameter to the find function because I've already specified what user I'm using via the model chain.

Any ideas why this is not working? The sql statment it runs is simply a SELECT on Groups WHERE 1 = 1 (so not filtering at all).

share|improve this question

add comment

1 Answer

up vote 0 down vote accepted

No, the object User really acts more like a class than a object. You can say Cake doesn't fully implement Active Record pattern (I think Cake 3.0 may fix that, not sure). So yes, you still need to set the condition for the find. And you don't have to set($this->Auth->user);

You are probably not very familiar with Cake: in Cake, you hardly ever have to instantiate Model objects at all. They are created for you based on the current controller and model relationships that you specify.

share|improve this answer

OK thanks for your answer. I am still learning cake, yes. The only reason I had that variable instantiated is because I wanted to use it in this manner - as that is not possible then you're right - I have no use for it. – MikeMurko Aug 26 '11 at 19:21

add comment

Your Answer

Subscribe English

look up any word, like fuck:

Scapulation dumping refers to the process of dumping scapula’s. Little is known about Scapula’s (from the latin word Scapulaiincredabilous) except that they are found at incredible depths within the worlds oceans. It is within their home environment that they creep and crawl over the ocean floors gathering the bubbles from shark farts.

Deep sea trawlers net the scapulas, extract the shark farts and encapsulate them in a vat of spirits which are ultimately sold to the manufacturers of spirit levels and used for the bubbles found within those levels.

Extreme caution must be applied throughout the action to avoid the shark farts entering the seamen’s blood stream and causing “the bends”.

The Scapula’s are not harmed during the process and are dumped back into the sea once their valuable cargo’s are extracted hence the expression “scapulation dumping”.

Prof J.S. Miller

Scapulation Dumping refers to dumping scapula's and is associated with production of Pioneer Brickies Levels

by Ormbo January 22, 2008

2 3

Words related to Scapulation Dumping refers to dumping scapula's:

dumping farts levels scapula sharks

Take the 2-minute tour ×

This seems like an ideal place to ask a question that has been keeping me confused for a little while now. But I apologize if I've posted in the wrong area.

My question is that I don't seem to understand the temperature trend that I am seeing from my temperature sensors. I'll start of with my setup: Essentially, I have a metal box with two temperature sensors. 1) Is mounted right at the base of the box [Temp Sensor #1] 2) The second temperature sensor is mounted 2 inches from the top of the box [Temp Sensor #2]. The box is about 6 inches in height.

I've placed a heating pad at the base of the box (Temp Sensor #1 lies right in the center of this pad. A graphic showing my setup:

I've also implemented a simple on/off temperature controller, that senses when the temperature goes above a certain set-point and turns off (hence you see the highs/lows for Temp Sensor #1).

As you can see from the image, the lower sensor (blue) has the peaks/troughs corresponding to when the heater turns on/off. The heater gets triggered every time the sensing temperature goes below the set point. What I don't understand is why the top sensor (red) has a periodically decreasing trend (it doesn't have highs/lows similar to the bottom sensor)? It doesn't seem to be affected by the heater turning on at all? Even though it is merely 4" away from the heater inside the metal enclosure.

I understand that Sensor#1 is probably changing immediately due to the heater very quickly affecting the metal base temperature via conduction. Whereas the second sensor is probably measuring the air around the metal enclosure at the top, and since air is an insulator, it takes longer to heat up. But there should be at-least some highs and lows I'd imagine. The continuous decreasing trend doesn't make any sense ...

Then, I suspected that perhaps my second temperature sensor was damaged. But that wasn't the case. I've tested both sensors and they work fine. Also here is a graph of the temperature trend, when I place the enclosure (with the sensors) in the freezer with no heater action. Intuitively as you can imagine, there is merely a decreasing trend for both sensors (shown below) due to the effect of the freezer:

Any suggestions please as to why I notice no temperature variation at Sensor #2 location when the heater turns on/off?

share|improve this question

Here is the trend with no heater and the freezer merely cooling: i47.tinypic.com/2430vp2.png , I couldn't post it in the original post since I do not have sufficient rep points. The heater pad type, I am using is this: winemakersdepot.com/Brewers-and-Wine-Making-Heat-Pad-P700.aspx – c0d3rz Jan 31 '13 at 0:32

I see when the heater is turned on an off, but when is the cooling activated? – mdma Jan 31 '13 at 1:32

What temp are you trying to stabilize at and is this metal enclosure insulated? Especially the lid. – brewchez Feb 2 '13 at 22:19

You've also inappropriately applied a linear fit to data that isn't behaving that way. You state that its a continuous decreasing trend. But it isn't, it has stabilize half way through the data collection period. – brewchez Feb 2 '13 at 22:24

add comment

1 Answer

up vote 1 down vote accepted

It's to do with thermal inertia. If you look closely at the graph you'll see there are highs and lows for sensor #2 also - just much smaller than sensor #1, and they have the same period (time interval) as sensor #1, indicating they stem from the same heating oscillation.

You're of course right when you say that air is a poor conductor, and so it will essentially dampen the effect of the heater - there are rises and falls, just over a smaller, dampened range.

I'm not entirely clear on when the freezer switches on and why it would be on when the heater is also on, but it seems that's the case from the graph. Another reason for the imbalance is that the freezer has a far greater cooling effect than the 25W heater can heat. Most freezers are in the order of 120W or more, and have a Coefficient of Performance (CoP) of 2 or more when chilling to beer temperature, so you're getting about 240W of real cooling power vs 25W of heating power (resistance heating has a CoP <= 1) - so a 10:1 difference.

I imagine the main reason you're seeing any fluctuation in sensor 2 at all is because of the box, which confines the convection of heated air to within the box. I imagine if you removed the box and did the same thing, you'd see very little change in sensor #2 because of the large volume of cooler air surrounding it.

share|improve this answer

Hi mdma: The freezer is always switched on, it is never turned off. Also, I'm using a 120W heater instead, so the heating/cooling difference is not that large I think. If I go with the idea of thermal inertia, why is that the lower sensor (Sensor #1) changes so quickly then? Shouldn't it face the same problem? – c0d3rz Jan 31 '13 at 2:56

The lower sensor changes quickly because the rate of conduction through contact with #1 is much quicker than conduction through the air to #2. – mdma Jan 31 '13 at 2:59

Hmm, just one last question about this CoP. Why is that the CoP is greater than 1 in the case of cooling? – c0d3rz Jan 31 '13 at 20:59

The CoP for compressor/inverter systems like fridges and heating pumps because they are just moving heat from one place to another, so the net heat output can be higher than the energy consumed. With a resistance based heat pad, this turns electrical energy into heat, and so the amount of heat produced can never be more than the electricity consumed. – mdma Jan 31 '13 at 21:09

add comment

Your Answer

• By

Rockstar Games

Image from L.A. Noire

Rockstar Games last week released L.A. Noire, a hypnotic detective thriller developed by Australian studio Team Bondi. The videogame has been called groundbreaking for the facial-recognition technology used to create its characters, and the resulting psychological dimensions of gameplay. In witness questioning and interrogation sequences, the player, as young detective Cole Phelps, tries to ascertain the truths and lies in what he’s told. Because L.A. Noire’s cast of characters feature the recorded facial performances of its actors, rather than hand-animated abstractions of feeling, the gameplay carries the emotional ambiguity and weight of a cop show or film noir. We spoke via email with Jeronimo Barrera, VP of product development at Rockstar, about the process of building the game.

Speakeasy: The team used maps from the Works Progress Administration, archival photographs from newspapers, and aerial photos as reference material for the environment. Can the eight-square-mile section of Los Angeles recreated in the game be considered documentary or used as research material in its own right?

Jeronimo Barrera: We went to a great deal of effort in every area, from the details of the interiors of certain key buildings, licensing classic billboards, and recreating former landmarks, but it’s not a block-for-block recreation—we are trying to make something that feels real, not that replicates reality. We have to make some concessions to game design, alongside more pedestrian things like building licensing. A great example of the lighter side of the issues involved is the fact that in 1947, all the palm trees in L.A. were still roughly head-high. But when they were designed that way in the game, they had the opposite effect to being realistic—they actually distracted players from the experience, so we designed them at full height. We also designed the game to be slightly less smoggy than L.A. was at the time.