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Your lambda is executed at some time after your loop has run completely. This means that the lambda will always be executed with the last object of the for loop.

To prevent this from happening, you can use a closure. Python has a simple way to create closures: Instead of a lambda use functools.partial

You could do the following to get the first 100 or so elements.

A common error:

Most common rand() implementations at least generate a number in the range [0...1), which is really all you need. You can scale a random number in one range to a number in another using the techniques outlined in the answers to this question, eg:

NewValue = (((OldValue - OldMin) * (NewMax - NewMin)) / (OldMax - OldMin)) + NewMin

For bash you have two choices: integer arithmetic or use a different tool.

Some of your choices for tools that support float arithmetic from the command-line include:

• a different shell (eg, zsh)
• perl: my $x = $minimum + rand($maximum - $minimum);
• ruby: x = min + rand * (max-min)
• awk: awk -v min=3 -v max=17 'BEGIN{srand(); print min+rand()*int(1000*(max-min)+1)/1000}'
  note: The original answer this was copied from is broken; the above is a slight modification to help correct the problem.
• bc: printf '%s\n' $(echo "scale=8; $RANDOM/32768" | bc )

... to name a few.

Just drop the last map at the end. The function is returning a list and your last map function is trying to take the first element of a list.

You define all functions inside of contractDate(), but you never call them and never assign them parameters.

Also:

Your return 0; statement should be placed after the for loop - otherwise after the first iteration 0 will be returned if the if condition is not fullfilled. Returning means that the function will halted before the iteration is complete.

Working sample:

Package author here. Yes, it would be wasteful to use more hashes/bands than you need. (Though keep in mind we are talking about kilobytes here, which could be much smaller than the original documents.)

The question is, what do you need? If you need to find only matches that are close to identical (i.e., with a Jaccard score close to 1.0), then you don't need a particularly sensitive search. If, however, you need to reliable detect potential matches that only share a partial overlap (i.e., with a Jaccard score that is closer to 0), then you need more hashes/bands.

Since you've read MMD, you can look up the equation there. But there are two functions in the package, documented here, which can help you calculate how many hashes/bands you need. lsh_threshold() will calculate the threshold Jaccard score that will be detected; while lsh_probability() will tell you how likely it is that a pair of documents with a given Jaccard score will be detected. Play around with those two functions until you get the number of hashes/bands that is optimal for your search problem.

After closely examining the source code, I found that indeed the Local Self Attention attends to the sequentially near tokens.

One thing that you can try is to use a switch point that is 1e-3 (or a certain minimum used) instead of zero. When the switch point is at zero and the condition is 1e-10 then the output will be 1 because it is greater than the switch point. This is needed because Gekko uses gradient based optimizers that have a solution tolerance of 1e-6 (default) so a solution within that tolerance is acceptable.

There are a couple examples in the documentation that may also help. You may also want to look at the sign2/sign3 functions and the max2/max3 functions that may also give you the desired result.

if2 Documentation

IF conditional with complementarity constraint switch variable. The traditional method for IF statements is not continuously differentiable and can cause a gradient-based optimizer to fail to converge. The if2 method uses a binary switching variable to determine whether y=x1 (when condition<0) or y=x2 (when condition>=0):

if3 Documentation

IF conditional with a binary switch variable. The traditional method for IF statements is not continuously differentiable and can cause a gradient-based optimizer to fail to converge. The if3 method uses a binary switching variable to determine whether y=x1 (when condition<0) or y=x2 (when condition>=0).

Usage

y = m.if3(condition,x1,x2)

Inputs:

• condition: GEKKO variable, parameter, or expression
• x1 and x2: GEKKO variable, parameter, or expression

Output:

• y = x1 when condition<0
• y = x2 when condition>=0