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\begin{array}{*{20}l} J(H) = tr(A^{T}DA) - 2tr\left(YHW^{T}X^{T}DA\right)\\ \quad +\lambda_{1} tr\left(W^{T}PW\right)+ \lambda_{2} tr\left(H^{T}QH\right) \\ \quad + tr\left(H^{T}Y^{T}YHW^{T}X^{T}DXW\right) \end{array}
(27)
\begin{array}{@{}[email protected]{}} tr\left(H^{T}\Lambda HB\right) \leq \sum_{i}\sum_{j}\left(\Lambda H^{\prime} B\right)_{{ji}}\frac{H^{2}_{{ij}}}{H^{\prime}_{{ij}}}, \end{array}
(28)

where, Λ , B , H are non-negative matrices, and Λ , B are symmetric matrices. And obviously the equality holds in Eq. when H = H .

In Eq. , if we do the substitutions: Λ = Y T Y , B = W T X T D X W , H = H , H = H , we see that the fifth term of Eq. is smaller than the fifth term of Eq. . However, the equality holds when H = H . Thus G ( H , H ) in Eq. is an auxiliary function of J ( H ). □

\begin{array}{*{20}l} \frac{\partial f(H)}{\partial H_{{ij}}} = -2\left(Y^{T}A^{T}DXW\right)_{{ij}}+2\lambda_{2}(QH)_{{ij}}\\ + 2\frac{\left(Y^{T}YH^{\prime}W^{T}X^{T}DXW\right)_{{ij}}H_{{ij}}}{{H^{\prime}}_{{ij}}} \end{array}
(29)
\begin{array}{*{20}l} \frac{\partial^{2} f(H)}{\partial H_{{ij}}\partial H_{{kl}}} = 2(Q)_{{jl}}\delta_{{ik}}\delta_{k\beta}\\ \quad+ \left(2\frac{\left(Y^{T}DYH^{\prime}W^{T}X^{T}XW\right)_{{ij}}}{H^{\prime}_{{ij}}} \right)\delta_{{jl}}\delta_{{ik}} \end{array}
(30)
\begin{array}{@{}[email protected]{}} H_{{ij}} = H^{\prime}_{{ij}}\frac{\left(Y^{T}A^{T}DXW\right)_{{ij}}}{\left(Y^{T}YH^{\prime}W^{T}X^{T}DXW + \lambda_{2} QH\right)_{{ij}}} \end{array}
(31)

By replacing H ( t +1) = H and H ( t ) = H , we would obtain the H update rule in Algorithm 3. Therefore, under this rule, the objective function J ( H ) of Eq. decreases monotonically, and hence completes the proof.

\begin{array}{*{20}l} tr\left(A-XW{H^{(t)}}^{T}Y^{T}\right)^{T}D\left(A-XW{H^{(t)}}^{T}Y^{T}\right)\\ \quad +\lambda_{1} tr\left(W^{T}PW\right) + \lambda_{2} tr\left({H^{(t)}}^{T}QH^{(t)}\right)\\ =\sum_{i=1}^{M}\sum_{j=1}^{N}\left(A-XW{H^{(t)}}^{T}Y^{T}\right)_{{ij}}D_{{ii}}\\ \quad +\lambda_{1} tr\left(W^{T}PW\right)+\lambda_{2}\sum_{k=1}^{n}\sum_{l=1}^{r} {H^{(t)}_{{kl}}}^{2}Q_{{kk}}\\ =\sum_{i=1}^{M}\left\|A_{i} - \left(XW{H^{(t)}}^{T}Y^{T}\right)_{i}\right\|^{2}D_{{ii}}\\ \quad +\lambda_{1} tr(W^{T}PW) + \lambda_{2} \sum_{k=1}^{n}\left\|H^{(t)}_{k}\right\|^{2} Q_{{kk}} \end{array}
\begin{array}{*{20}l} tr\left(A-XW{H^{(t+1)}}^{T}Y^{T}\right)^{T}D\left(A-XW{H^{(t+1)}}^{T}Y^{T}\right)\\ \quad +\lambda_{1} tr\left(W^{T}PW\right) + \lambda_{2} tr\left({H^{(t+1)}}^{T}QH^{(t+1)}\right)\\ =\sum_{i=1}^{M}\left\|A_{i} - \left(XW{H^{(t+1)}}^{T}Y^{T}\right)_{i}\right\|^{2}D_{{ii}}\\ \quad +\lambda_{1} tr\left(W^{T}PW\right) + \lambda_{2} \sum_{k=1}^{n}\left\|H^{(t+1)}_{k}\right\|^{2} Q_{{kk}} \end{array}
\begin{array}{*{20}l} {}r.h.s = \frac{1}{2}\sum_{i=1}^{M}\left(\left\|A_{i}-\left(XW{H^{(t+1)}}^{T}Y^{T}\right)_{i}\right\|^{2}\right.\\ \quad \left. -\left\|A_{i}-\left(\!XW{H^{(t)}}^{T}Y^{T}\!\right)_{i}\right\|\right)D_{{ii}} + \lambda_{2} \sum_{k=1}^{n}\left(\! \left\|H^{(t+1)}_{k}\right\|^{2} \right.\\ \quad \left.- \|H^{(t)}_{k}\|^{2}\right)Q_{{kk}}\\ =\frac{1}{2}\sum_{i=1}^{M}\left(\left\|A_{i}-\left(XW{H^{(t+1)}}^{T}Y^{T}\right)_{i}\right\|^{2} D_{{ii}} - \frac{1}{D_{{ii}}}\right)\\ \quad +\lambda_{2} \sum_{k=1}^{n}\left(\left\|H^{(t+1)}_{k}\right\|^{2}Q_{{kk}} - \frac{1}{Q_{{kk}}}\right) \end{array}
\begin{array}{*{20}l} l.h.s =\sum_{i=1}^{M}\left(\sqrt{\left\|A_{i}-\left(XW{H^{(t+1)}}^{T}Y^{T}\right)_{i}\right\|^{2} }\right.\\ \quad \left. - \sqrt{\left\|A_{i}-\left(XW{H^{(t)}}^{T}Y^{T}\right)_{i}\right\|^{2}}\right)\\ \quad +\lambda_{2}\sum_{k=1}^{n}\left(\sqrt{\left\|H^{(t+1)}_{k}\right\|^{2}}-\sqrt{\left\|H^{(t)}_{k}\right\|^{2}}\right)\\ =\sum_{i=1}^{M}\left(\left\|A_{i}-\left(XW{H^{(t+1)}}^{T}Y^{T}\right)_{i}\right\| \right.\\ \quad \left.-\left\|A_{i}-\left(XW{H^{(t)}}^{T}Y^{T}\right)_{i}\right\|\right)\\ \quad +\lambda_{2}\sum_{k=1}^{n}\left(\sqrt{\left\|H^{(t+1)}_{k}\right\|^{2}}-\sqrt{\left\|H^{(t)}_{k}\right\|^{2}}\right)\\ = \sum_{i=1}^{M}\left(\left\|A_{i}-\left(XW{H^{(t+1)}}^{T}Y^{T}\right)_{i}\right\| - \frac{1}{D_{{ii}}}\right)\\ \quad +\lambda_{2} \left(\sum_{k=1}^{n}\left\|H^{(t+1)}_{k}\right\| - \frac{1}{Q_{{kk}}}\right) \end{array}
\begin{array}{*{20}l} l.h.s - r.h.s = \sum_{i=1}^{M}\left(\left\|A_{i}-\left(XW{H^{(t+1)}}^{T}Y^{T}\right)_{i}\right\|\right.\\ \quad \left.-\left\|A_{i}-\left(XW{H^{(t+1)}}^{T}Y^{T}\right)_{i}\right\|^{2}\frac{D_{{ii}}}{2} - \frac{1}{2D_{{ii}}}\right)\\ \quad +\lambda_{2}\sum_{k=1}^{n}\left(\left\|H^{(t+1)}_{k}\right\|-\left\|H^{(t+1)}_{k}\right\|^{2}\frac{Q_{{kk}}}{2}-\frac{1}{2Q_{{kk}}}\right)\\ =\sum_{i=1}^{M}\frac{D_{{ii}}}{2}\left(\frac{\left\|A_{i}-\left(XW{H^{(t+1)}}^{T}Y^{T}\right)_{i}\right\|}{D_{{ii}}}\right.\\ \quad \left.-\left\|A_{i}-\left(XW{H^{(t+1)}}^{T}Y^{T}\right)_{i}\right\|^{2}-\frac{1}{D^{2}_{{ii}}}\right)\\ \quad +\lambda_{2}\sum_{k=1}^{n}\frac{Q_{{kk}}}{2}\left(\frac{\left\|H^{(t+1)}_{k}\right\|}{Q_{{kk}}}-\left\|H^{(t+1)}_{k}\right\|^{2}-\frac{1}{Q^{2}_{{kk}}}\right)\\ =\sum_{i=1}^{M}\frac{(-D_{{ii}})}{2}\left(\left\|A_{i}-\left(XW{H^{(t+1)}}^{T}Y^{T}\right)_{i}\right\|-\frac{1}{D_{{ii}}} \right)^{2}\\ \quad +\lambda_{2}\sum_{k=1}^{n}\frac{(-Q_{{kk}})}{2}\left(\left\|H^{(t+1)}_{k}\right\|-\frac{1}{Q_{{kk}}}\right)^{2}\\ \leq 0\end{array}

The above inequality holds because, D , Q are non-negative matrices, and the sum of non-positive numbers is always non-positive. This completes the proof. □

Tip 3: Check the property existence

Fortunately, JavaScript offers a bunch of ways to determine if the object has a specific property:

My recommendation is to use  in  operator. It has a short and sweet syntax.  in  operator presence suggests a clear intent of checking whether an object has a specific property, without accessing the actual property value.

 obj.hasOwnProperty('prop')  is a nice solution too. It's slightly longer than  in  operator and verifies only in object's own properties.

The 2 ways that involve comparing with  undefined  might work... But it seems to me that  obj.prop !== undefined  and  typeof obj.prop !== 'undefined'  look verbose and weird, and expose to a suspicions path of dealing directly with  undefined  .

Let's improve  append(array, toAppend)  function using  in  operator:

 'first' in toAppend  (and  'last' in toAppend  ) is  true  whether the corresponding property exists,  false  otherwise.

The usage of  in  operator fixes the problem with inserting falsy elements  0  and  false  . Now, adding these elements at the beginning and at the end of  [10]  produces the expected result  [0, 10, false]  .

Tip 4: Destructuring to access object properties

When accessing an object property, sometimes it's necessary to indicate a default value if the property does not exist.

You might use  in  accompanied with ternary operator to accomplish that:

The usage of ternary operator syntax becomes daunting when the number of properties to check increases. For each property you have to create a new line of code to handle the defaults, increasing an ugly wall of similar looking ternary operators.

In order to use a more elegant approach, let's get familiar with a great ES2015 feature called object destructuring .

allows inline extraction of object property values directly into variables, and setting a default value if the property does not exist. A convenient syntax to avoid dealing directly with  undefined  .

Indeed, the property extraction now looks short and meaningful:

To see things in action, let's define an useful function that wraps a string in quotes.  quote(subject, config)  accepts the first argument as the string to be wrapped. The second argument  config  is an object with the properties:

Applying the benefits of the object destructuring, let's implement  quote()  :

 const { char = '"', skipIfQuoted = true } = config  destructuring assignment in one line extracts the properties  char  and  skipIfQuoted  from  config  object. If some properties are not available in  config  object, the destructuring assignment sets the default values:  '"'  for  char  and  false  for  skipIfQuoted  .

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By Christoph Meyer

Uber and Lyft may say otherwise, but ridehailing is inching closer to personalized pricing: the ability to charge the maximum price -- known in economics as the reservation price -- but enable that to be different for every customer, for an identical good or service. Considered one of the most elusive targets of consumer economics, personalized pricing is difficult to achieve. Even airlines, which have become savvy at price discrimination ( Kenneth Cole New York Mens High Chair Brown 8G38dxe0dO
) can’t compete with ridehailers. Here’s why: Uber and Lyft have far more data points (people take many more rides than flights) and do not have price aggregators (Google Flights, Expedia, etc.) that allow price comparisons.

As ride-hailing giants collect more and more data and perfect their algorithms, they will simultaneously enhance their ability to extract as much value from each user as possible.

Until now, Uber and Lyft have focused on riders’ locations and demand (e.g. time) to vary the price it charges and to trigger surge pricing when demand is high. But these companies have the ability to go far beyond that, testing and eventually mastering the ability to price rides based on numerous other factors that could be considered indicators of a given individual’s willingness to pay. Some of these may include:

What could this look like? Imagine a young professional with a San Francisco area code who hails a vehicle from her iPhone. It’s late at night and her phone charge is down to under 10%. Since she usually takes Uber or Lyft for her job, she has a selective AmEx on file for payment. She also has a tendency to accept surge pricing, since she often travels at rush hour and can expense travel to her client. The company has found a customer that shows signs of being more inclined to pay more and can shift its price accordingly.

But how is a company able to do this? There are two reasons:

1) Ridehailers know the upper limit of your pricing comfort

Uber and Lyft have an incredibly rich dataset for each individual rider. With riders taking an average of 4 rides per month (back in 2014), these companies have built up a detailed history of each rider that will only continue to grow. Particularly on trips with same starting and ending points, they can see how much a particular rider was willing to pay in the past. Having introduced upfront pricing, companies can even test rider’s willingness to pay in real time by seeing whether riders hail the ride or close the app and try again.

2) Ridehailers can charge different prices for the same service

Uber and Lyft provide a customized and personal service to each user. By having the price offered on an individual smartphone, the company can charge different prices to each rider without other riders knowing. Even if they would know, it would likely be irrelevant: few people have identical starting and ending locations. Ridehailing can not only know the reservation price of individuals but also charge them this exact price in private.

While companies have caused outrage in the past with surge/prime-time pricing, they hold even greater power going forward. Having moved to partially mask surge pricing and continuing to collect greater amounts of data on its customers, these transportation companies have the potential to achieve one of economics/business’ holy grails: perfect price discrimination.

The results of perfect price discrimination are mixed across parties. Uber and Lyft stand to gain by increasing prices and thus revenues. Drivers also stand to gain from higher fares from riders. Riders, on the other hand, are those that will be most negatively affected. As Uber and Lyft continue to take share of the market, riders will have fewer choices and thus fewer alternatives to paying these higher fares. Regulation may ultimately be needed to provide pricing transparency to riders and ensure that people are being fairly charged for similar services.

Phone type : iPhone users tend to be wealthier than Android users, meaning companies could charge them a higher price

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