enter image description herewhen i try to make an if statment in the on Pressed an error occured which is => 'Equality operator '==' invocation with references of unrelated typesthis is my code and in the text you will see my list i want to make an if statment in onPressed method
final List<Movie> movies = [
Movie(
imageUrl: 'assets/images/5s.PNG',
title: 'video 1',
categories: 'Fantasy, Sci-fi',
year: 2018,
country: 'uk',
length: 60,
description:
'Our friendly neighborhood Super Hero decides to join his best friends Ned, MJ, and the rest of the gang on a European vacation. However, Peter\'s plan to leave super heroics behind for a few weeks are quickly scrapped when he begrudgingly agrees to help Nick Fury uncover the mystery of several elemental creature attacks, creating havoc across the continent.',
screenshots: [
'assets/images/1.PNG',
'assets/images/2.PNG',
'assets/images/3.PNG',
],
),
Movie(
imageUrl: 'assets/images/Star.PNG',
title: 'Star',
categories: 'Adventure, Family, Fantasy',
year: 2000,
country: 'US',
length: 100,
description:
'All Clara wants is a key - a one-of-a-kind key that will unlock a box that holds a priceless gift from her late mother. A golden thread, presented to her at godfather Drosselmeyer\'s annual holiday party, leads her to the coveted key-which promptly disappears into a strange and mysterious parallel world. It\'s there that Clara encounters a soldier named Phillip, a gang of mice and the regents who preside over three Realms: Land of Snowflakes, Land of Flowers, and Land of Sweets. Clara and Phillip must brave the ominous Fourth Realm, home to the tyrant Mother Ginger, to retrieve Clara\'s key and hopefully return harmony to the unstable world.',
screenshots: [
'assets/images/l1.PNG',
'assets/images/nl2.PNG',
'assets/images/kl3.PNG',
],
),
Movie(
imageUrl: 'assets/images/johnyjohny.PNG',
title: 'Johny',
categories: 'Adventure, Fantasy',
year: 2019,
country: 'ire',
length: 65,
description:
'Woody, Buzz Lightyear and the rest of the gang embark on a road trip with Bonnie and a new toy named Forky. The adventurous journey turns into an unexpected reunion as Woody\'s slight detour leads him to his long-lost friend Bo Peep. As Woody and Bo discuss the old days, they soon start to realize that they\'re two worlds apart when it comes to what they want from life as a toy.',
screenshots: [
'assets/images/johny.PNG',
'assets/images/johny.PNG',
'assets/images/johny.PNG',
],
),
];`
You get this error because you try to compare different types with each other; movies(List) and movies[0](a member of movies whose type is Movie) and they can not be equal never. For what reason you want to use "if" statement if you explain that, we may help you easier.
enter image description here
[ ][2]
these are two images showing my code , i want when the user click on the arrow-play button to take him to other page where the video is displayed depending on the list of movies
Related
In Excel I have a data table of Paired Items that are tagged with an identifier. Essentially, named linkages.
Worksheet: Links
Tag
Point-A
Point-B
Route 1
Home
Office
Route 2
Home
Grocery 1
Happy Hour
Office
Bar
Sad Hour
Office
Dump
Headaches
Bar
Pharmacy
Sick
Bar
Dump
Route 3
Office
Moms
Route 4
Office
Park
Victory
Park
Bar
Discard
Park
Dump
I want to transform this data into a grid of all points in rows and columns with the tag placed at the intersection (Much like old paper road maps with grids for city distances)
Worksheet: Grid
A \ B
Bar
Dump
Grocery 1
Home
Home
Moms
Office
Office
Park
Pharmacy
Bar
Sick
Happy Hour
Victory
Headaches
Dump
Sick
Sad Hour
Discard
Grocery 1
Route 2
Home
Route 1
Home
Route 2
Moms
Route 3
Office
Happy Hour
Sad Hour
Route 1
Route 3
Office
Route 4
Park
Victory
Discard
Route 4
Pharmacy
Headaches
I have written the following M code for transforming, but it seems a bit wayward and overwrought. I am using bit coding of points to construct a join key, so the bitting process will probably break around 32 points.
Is there a shorter set of LETs that do the same transform to grid ?
Is there a way to create a key that is Min(Point-A,Point-B) delimited concatenation with Max(Point-A,Point-B), and thus not rely of bitting?
M code (copied from Advanced Editor)
let
LinksTable = Table.SelectRows(Excel.CurrentWorkbook(), each [Name] = "Links"),
Links = Table.RemoveColumns(Table.ExpandTableColumn(LinksTable, "Content", {"Tag", "Point-A", "Point-B"}), "Name"),
AllPoints = Table.Combine(
{ Table.SelectColumns(Table.RenameColumns(Links,{"Point-A", "Point"}), "Point"),
Table.SelectColumns(Table.RenameColumns(Links,{"Point-B", "Point"}), "Point")
}),
ThePoints = Table.Sort(Table.Distinct(AllPoints),{"Point"}),
PointsIndexed = Table.AddIndexColumn(ThePoints, "Index", 0, 1, Int64.Type),
PointsBitted = Table.RemoveColumns(Table.AddColumn(PointsIndexed, "Bit", each Number.Power(2, [Index]), Int64.Type),"Index"),
AllPairsBitted = Table.Join(
Table.RenameColumns(PointsBitted, {{"Point", "Point-A"}, {"Bit", "Bit-A"}}), {},
Table.RenameColumns(PointsBitted, {{"Point", "Point-B"}, {"Bit", "Bit-B"}}), {},
JoinKind.FullOuter
),
AllPairsKeyed = Table.RemoveColumns(
Table.AddColumn(AllPairsBitted, "BitKeyPair", each Number.BitwiseOr([#"Bit-A"],[#"Bit-B"])),
{ "Bit-A", "Bit-B"}
),
#"Links-A-Bitted" = Table.Join(
Links, "Point-A",
Table.RenameColumns(PointsBitted,{{"Point", "Point-A"}, {"Bit", "Bit-A"}}), "Point-A"
),
#"Links-AB-Bitted" = Table.Join(
#"Links-A-Bitted", "Point-B",
Table.RenameColumns(PointsBitted,{{"Point", "Point-B"}, {"Bit", "Bit-B"}}), "Point-B"
),
LinksKeyed = Table.RemoveColumns(
Table.AddColumn(#"Links-AB-Bitted", "BitKeyLink", each Number.BitwiseOr([#"Bit-A"],[#"Bit-B"])),
{ "Bit-A", "Bit-B"}
),
AllPairsTagged = Table.Sort( Table.RemoveColumns(
Table.Join(
AllPairsKeyed, "BitKeyPair",
Table.SelectColumns(LinksKeyed, {"BitKeyLink", "Tag"}), "BitKeyLink",
JoinKind.LeftOuter
),
{"BitKeyPair", "BitKeyLink"}
),
{"Point-A", "Point-B"}
),
Grid = Table.Pivot(AllPairsTagged, List.Distinct(AllPairsTagged[#"Point-B"]), "Point-B", "Tag", List.First)
in
Grid
I think you can use PIVOT to achieve this. Using directly this functionality would not work because you are looking for symmetry of columns and rows.
The trick is to force that symmetry, appending values from Point-B into values of Point-A.
Steps
Create a secondary table and reorder the columns in the opposite way that the original table, so Tag, Point-B and Point-A.
On the secondary table, rename the columns to Tag, Point-A and Point-B in that order. Append usually take column names literally, so without renaming it would append the names of the same columns.
Pivot on column Point-B without aggregating data.
Reorder the columns using Point-A as a reference, so you have symmetry of columns and rows.
It's worth mentioning that's good practice to Buffer the source table because is used multiple times across the calculations.
Calculation
let
Source = Table.FromRows(Json.Document(Binary.Decompress(Binary.FromText("Zc69CsMgEMDxVwnOLv14ghKoS2looEvIcJwXIkEMZxx8+6axiUInwd/99bpOvFxYqDoJKZSztB7PYTBIope7nbPd2SFxXMe/rGCeY6Vc4JxJcQPetAX9Z3Wwc0oJNOBI/hdI0YzAFjCm1uB0yBGldS7lgw9nfWHX0hrgabO3wcVx3K/yirXxCKwzpK/6Dw==", BinaryEncoding.Base64), Compression.Deflate)), let _t = ((type nullable text) meta [Serialized.Text = true]) in type table [Tag = _t, #"Point-A" = _t, #"Point-B" = _t]),
BufferedSource = Table.Buffer(Source),
SecondTable = Table.ReorderColumns(BufferedSource,{"Tag","Point-B","Point-A"}),
SecondTableRenameCols = Table.RenameColumns(SecondTable,{{"Point-A","Point-B"},{"Point-B","Point-A"}}),
AppendTables = Table.Combine({BufferedSource,SecondTableRenameCols}),
PivotTables = Table.Pivot(AppendTables, List.Distinct(AppendTables[#"Point-B"]), "Point-B", "Tag"),
ReorderCols = Table.ReorderColumns( PivotTables, PivotTables[#"Point-A"])
in ReorderCols
Output
Point-A
Bar
Dump
Grocery 1
Home
Moms
Office
Park
Pharmacy
Bar
Sick
Happy Hour
Victory
Headaches
Dump
Sick
Sad Hour
Discard
Grocery 1
Route 2
Home
Route 2
Route 1
Moms
Route 3
Office
Happy Hour
Sad Hour
Route 1
Route 3
Route 4
Park
Victory
Discard
Route 4
Pharmacy
Headaches
In a survey of 100 people, I am asking each person to choose between product A and product B. I ask each person this question 3 times, but each time I present a different set of products. Say, first time, Person 1 is asked to choose between 'Phone 1' and 'Phone 2', given certain attributes of each phone. The second time the choice is again 'Phone 1' vs. 'Phone 2', but a different set of attributes for each phone.
A person is presented three attributes associated with the two phone alternatives every time the question is asked. So, each time between Phone 1 and Phone 2, the attributes of the phone such as cost, memory and camera pixels are presented so that user can choose which set of attributes is most attractive, Phone 1's or Phone 2's.
Overall, 3*100 = 300 responses; 3 responses per person. Each time the attributes cost, memory and camera pixels presented and user asked to choose the feature set they prefer.
My goal is to analyze how users value features of a phone vs. cost of the phone.
In this scenario, can I use a MNL - even though each time I asked the person a question, I only presented two choices ? My understanding is that MNL is sued when (a) there are multiple choices and (b) the choice options do not change across observations, i.e. each person is asked to choose between multiple products, say A, B, C and A, B, C do not change across observations.
In the scenario described above, the two choices varied across the three times the same person was asked the question ? If not MNL, should I rather create a binary logit model given that user only had to choose between two options when the question was asked (even though he was asked the question three times)? If I can use binary logit, should I be concerned that the choice set of products change across observations ? or should I let the attributes defined in each of the rows address the differences in product choices across observations.
I have setup the data as follows (thinking I can do MNL but may be I should set it up differently and use another modeling approach?):
I am working on designing and analyzing similar survey but mine is related to transportation. I am at the beginning level and I am still new to the whole concept, however I will give you an advice and reference maybe it is helpful.
First point: I have come cross 3 models as following from a useful video on YouTube:
MNL refers to Multinomial Logit Model. MNL is used with
alternative-invariant regressors (for example salary of participant
in the survey, or his/her gender …).
Conditional logit model is used with alternative-invariant (gender,
salary, education level …) and alternative-variant regressors (cost
of the product, memory, camera pixel …)
Mixed logit model which uses random parameters. It is also used with
alternative-invariant (gender, salary, education level …) and
alternative-variant regressors (cost of the product, memory, camera
pixel …)
Note regarding alternative-invariant and alternative-variant regressors:
The gender of person participating in the survey will NOT vary between Product A or Product P, so it is alternative-invariant regressor. While price of product could vary between Product A and Product B so it is called alternative-variant regressors.
Based on above I assume you need to use conditional logit model or mixed logit model.
For me I couldn’t find a special function in R for the conditional logit model or mixed logit model. The same mlogit function is used, refer to the examples below for the help of mlogit package:
a pure "multinomial model"
summary(mlogit(mode ~ 0 | income, data = Fish))
a pure "conditional" model
summary(mlogit(mode ~ price + catch, data = Fish))
a "mixed" model
m <- mlogit(mode ~ price+ catch | income, data = Fish)
summary(m)
same model with charter as the reference level
m <- mlogit(mode ~ price+ catch | income, data = Fish, reflevel = "charter")
From the examples above, I think (but NOT sure) that in the Manual of mlogit package, they refer to mixed logit when you used both alternative-invariant and alternative-variant regressors. While conditional model when you have only alternative-variant regressors. On the other hand, multinomial model when you have only multinomial alternative-invariant regressors.
Second point: There is something called “panel data” when you are asking the same person to choose one product for each choice-set. Same person here means that in your model you are taking into consideration, the gender, the salary, the education level … which they will stay the same for the same person. Check this: https://en.wikipedia.org/wiki/Panel_data
To use panel techniques please refer to help in mlogit package in R. I am quoting from it the following:
“panel only relevant if rpar is not NULL and if the data are repeated observations of the same unit ; if TRUE, the mixed-logit model is estimated using panel techniques”
So in my understanding, if you want to use the panel techniques you have to use random draws because panel will be true and rpar will not be NULL.
Moreover, for example about using the panel data, please refer to the below example from “Estimation of multinomial logit models in R : The mlogit Packages” by Yves Croissant
data("Train", package = "mlogit")
Tr <- mlogit.data(Train, shape = "wide", varying = 4:11, choice = "choice", sep = "_", opposite = c("price", "time", "change", "comfort"), alt.levels=c("A", "B"), id.var ="id")
Train.ml <- mlogit(choice ~ price + time + change + comfort, Tr)
Train.mxlc <- mlogit(choice ~ price + time + change + comfort, Tr, panel = TRUE, rpar = c(time = "cn", change = "n", comfort = "ln"), correlation = TRUE, R = 100, halton = NA)
Train.mxlu <- update(Train.mxlc, correlation = FALSE)
I hope that is useful to you.
I would like to have ActiveRecord objects to run tests against that span several days.
For example, I want x number of Posts over the last x days so that I can calculate the average rating for posts posted yesterday. This way I know it didn't include tests from posts in the previous days.
Would I have to seed the database to do this or am I writing bad tests? :)
Take the following simplified example where your posts table has the rating column and a created_at column.
You can 'seed'/setup your test to mimic this, without needing to actually seed your development database for this.
let(:post_one) { Post.create(rating: 5, created_at: Time.now) }
let(:post_two) { Post.create(rating: 5, created_at: Time.yesterday) }
it "calculates the average correctly" do
expect "your expectation"
end
Ideally make use of FactoryGirl rather than simply Post.create. You can then switch that out for FactoryGirl.create(:post) { etc... }
Say I have a list of PROLOG facts, which are Blog Post titles, the author, and the year they were written:
blogpost('Title 1', 'author1' 2012).
blogpost('Title 2', 'author1', 2011).
blogpost('Title 3', 'author1' 2010).
blogpost('Title 4', 'author1', 2006).
blogpost('Title 5', 'author2' 2009).
blogpost('Title 6', 'author2', 2011).
I want to write a rule which has two parameters/inputs, the author, and the year. If the author entered has written an article after the specified year, PROLOG would return true.
Here is what I have tried:
authoredAfter(X,Z) :-
blogpost(_,X,Z),
So if I queried ?- authoredAfter('author1',2010). PROLOG would return true because the Author has written an article in 2010. However, if I query ?- authoredAfter('author1',2009)., it would return false, but I want it to return true because author1 has written an article after that year.
My question is, how do I compare the user input value to the value in the fact?
You need two use two distinct variables for the year of the article and for the year you want to start your search from and compare them. Something like this:
authoredAfter(Author, Year1):-
blogpost(_, Author, Year2),
Year2 >= Year1.
You express the fact that Author authoredAfter Year1 if there is a blog post written by Author in Year2 and Year2 >= Year1.
If you issue a query to see if author1 wrote anything after 2009:
?- authoredAfter(author1, 2009).
true ;
true ;
true ;
false.
the goal is satisfied three times as author1 has 3 blog posts after 2009 (2010, 2011, 2012). If you want to get a single answer, no matter how many such articles exist, you can use once/1:
authoredAfter(Author, Year1):-
once((blogpost(_, Author, Year2),
Year2 >= Year1)).
?- authoredAfter(author1, 2009).
true.
I have an application where users can:
Write reviews about products
Add comments to products
Up / Down vote reviews
Up / Down vote comments
Every Up/Down vote is recorded in a db table.
What i want to do now is to create a ranking of the most active users in the last 4 weeks.
Of course good reviews should be weighted more than good comments. But also e.g. 10 good comments should be weighted more than just one good review.
Example:
// reviews created in recent 4 weeks
//format: [ upVoteCount, downVoteCount ]
var reviews = [ [120,23], [32,12], [12,0], [23,45] ];
// comments created in recent 4 weeks
// format: [ upVoteCount, downVoteCount ]
var comments = [ [1,2], [322,1], [0,0], [0,45] ];
// create weight vector
// format: [ reviewWeight, commentsWeight ]
var weight = [0.60, 0.40];
// signature: activties..., activityWeight
var userActivityScore = score(reviews, comments, weight);
... update user table ...
List<Users> users = "from users u order by u.userActivityScore desc";
How would a fair scoring function look like?
How could an implementation of the score() function look like? How to add a weight g to the function so that reviews are weighted heavier? How would such a function look like if, for example, votes for pictures would be added?
These kinds of algorithms can be quite complex. You might want to take a look into these books:
Collective Intelligence in Action, Satnam Alag, Manning, 2008, ISBN 1933988312
Algorithms of the Intelligent Web, Haralambos Marmanis, Manning 2009, ISBN 1933988665
Programming Collective Intelligence: Building Smart Web 2.0 Applications, Toby Segaran, O'Reilly Media 2007, ISBN 0596529325