I'm using the early ProjectEuler problems as a way to get to know Factor. Already in the very first problem I don't find a satisfactory solution.
I can solve the division test with this
: 3or5divisible ( n -- ? ) [ 3 mod ] [ 5 mod ] bi * 0 = ;
But what I don't like is the repetition of the mod. Of course it's only two times, but another problem might need to check 200.
I tried with map, curry, bi, 2bi, bi#, arrays and plain stack values etc. I always get a stack underflow or an effect mismatch (when using map). I have have not yet found a way to see the results of my trials in the inspector.
How can I factor that mod out and have it applied to { 3 5 } or an equivalent stack?
Cool would be two variants e.g. of a mod3and5 (including effect specification); one that leaves 2 1 on the stack for input 11 and one that returns { 2 1 }.
I can only think of doing it this way, which removes one of the mods, but ends up adding a bi-curry# function.
: 3or5divisible ( n -- ? ) 3 5 [ mod ] bi-curry# bi * 0 = ;
Related
In my code I use Julia's prod() function to do a product over the elements of a list. However, sometimes that list is empty, in which case I want prod(myList) to just return 0 (or any fixed number like 1). I tried searching online, but the only things I could find were for iterators or stuff like that.
I'm using Julia version 1.5.2.
Would a simple ternary operator work for your case?
isempty(my_list) ? 0 : prod(my_list)
What you want is incorrect/unconventional. The product of the elements of an empty sequence should be 1, because that is the multiplicative identity element.
"Any fixed number" is easy:
reduce(*, ls; init=1)
But this does not work well with zero, since that is an annihilator and sends the whole product to zero:
julia> ls = [1,2,3]
3-element Array{Int64,1}:
1
2
3
julia> reduce(*, ls; init=0)
0
Now, returning 1 and then checking for that works if you only have integers. It doesn't as soon as you have a product over rationals, since then the resulting 1 could also stem from x * (1/x).
julia> zeroprod(x) = isempty(x) ? zero(eltype(x)) : prod(x)
zeroprod (generic function with 1 method)
I'm am exploring methods of giving scores to different datapoints within a dataset. These points come from a mix of numbers and text string attributes looking for certain characteristics, e.g. if Col. A contains more than X number of "|", then give it a 1. If not, it gets a 0 for that category. I also have some that give the point when the value is >X.
I have been trying to do this with =IF, for example, =IF([sheet] = [Text], "1","0").
I can get it to give me 1 or 0, but I am unable to get a point total with sum.
I have tried changing the formatting of the text to both "number", "plain text", and have left it as automatic, but I can't get it to sum. Thoughts? Is there maybe a better way to do this?
FWIW - I'm trying to score based on about 12 factors.
Best,
Alex
The issue here might be that you're having the cell evaluate to either the string "0" or the string "1" rather than the number 0 or the number 1. That would explain why you're seeing the right things but the math isn't coming out right - the cell contents look like numbers, but they're really text, which the summation would then ignore.
One option would be to drop the quotation marks and write something like this:
=IF(condition, 1, 0)
This has the condition evaluate to 1 if it's true and 0 if it's false.
Alternatively, you could write something like this:
=(condition) * 1
This will take the boolean TRUE or FALSE returned by condition and convert it to either the numeric value 1 (true) or the numeric value 0 (false).
I got stuck with a specific question in R around concatenating columns of a data frame by using a wildcard. Perhaps I am searching wrongly. However I could not find a matching answer yet.
Here is my question:
I have a data frame df where each column represents a user (U1, U2, U3), e.g.:
> df <-data.frame(U1=1:3, U2=4:6, U3=7:9)
> df
> U1 U2 U3
1 1 4 7
2 2 5 8
3 3 6 9
I would like to concatenate the values from all users into a single vector as one would do using the c() function, e.g.:
> c(df$U1, df$U2, df$U3)
[1] 1 2 3 4 5 6 7 8 9
However, my number of users is large and varies over time. So, I look for an elegant dynamic way of concatenating the columns such as
> c(df$U*)
Unfortunately this does not seem to work. I played around with grep and regular expressions but could not get it to work. For sure, I could use a for-loop and program my own cat function but I assume there is a better way. I just don't find it. Maybe I am just blind. Hope you can help.
sub_df <- df[, grep(pattern ='^U.*', names(df))]
stack(df)$values
Hope this works for you. You could first subset some columns according to your need.
Coerce the data frame to a matrix first:
as.vector(as.matrix(df))
Use the bracket [ to select columns whose names match a certain expression:
df[, grep("U.*", colnames(df)), drop = FALSE]
I have a dataset of actions doing over time, an attribute 'Hour' ( contains values from 0 ->23 ). Now I want to create another attribute, say 'PartOfDay', which group 24 hours into 4 parts. For tuples have 'Hour' value of 0 to 5, then the 'PartOfDay' value should be 1; if 'Hour' value in [6,11], then the 'PartOfDay' value should be 2;...How can I do?
The codes would do this:
train['PartOfDay']=1
train.loc[(train.Hour>=6) & (train.hour<=11),'PartOfDay']=2
train.loc[(train.Hour>=12) & (train.hour<=17),'PartOfDay']=3
train.loc[(train.Hour>=18) & (train.hour<=23),'PartOfDay']=4
but it seems not so beautiful, I would like to know a more decent one if possible
Thank you for all your supports!!
While it is not clear what train.loc represents, a general approach to your problem is to use modulus function to set the RHS:
1 + int(train.Hour / 6)
We are given an integer N and we need to count the total number of permutations of numbers less than N. We are also given N-1 constraints. e.g.:
if N=4 then count permutations of 0,1,2,3 given:
0>1
0>2
0>3
I thought about making a graph and then counting total no of permutation of numbers at same level and multiply it with permutations at other level.e.g.:
For above example:
0
/ | \
/ | \
1 2 3 ------> 3!=6 So total no of permutations are 6.
But I have difficulty in implementing it in C++. Also, this question was asked in Facebook hacker cup, the competition is over now. I have seen code of other people and found that they did it using DFS. Any help?
The simplest way to do this is to use a standard permutation generator and filter out each permutation that violates the conditions. This is obviously very inefficient and for larger values of N is not computable. Doing this is sort of the "booby" option that these contests have which allows the less smart contestants to complete the problem.
The skilled approach requires insight into the ways of counting combinations and permutations. To illustrate the method I will use an example. Inputs:
N = 7
2 < 4
0 < 3
3 < 6
We first simplify this by combining the dependent conditions into a single condition, as follows:
2 < 4
0 < 3 < 6
Start with the longest condition, and determine the combination count of the gaps (this is the key insight). For example, some of the combinations are as follows:
XXXX036
XXX0X36
XXX03X6
XXX036X
XX0XX36
etc.
Now, you can see there are 4 gaps: ? 0 ? 3 ? 6 ?. We need to count the possible partitions of X's in these four gaps. The number of such partitions is (7 choose 3) = 35 (do you see why?). Now, we next multiply by the combinations of the next condition, which is 2 < 4 over the remaining blank spots (the Xs). We can multiply because this condition is fully independent of the 0<3<6 condition. This combination count is (4 choose 2) = 6. The final condition has 2 values in 2 spots = 2! = 2. Thus, the answer is 35 x 6 x 2 = 420.
Now, let's make it a little more complicated. Add the condition:
1 < 6
The way this changes the calculation is that before 036 had to appear in that order. But, now, we have three possible arrangements:
1036
0136
0316
Thus, the total count is now (7 choose 4) x 3 x (3 choose 2) = 35 x 3 x 3 = 315.
So, to recap, the procedure is you isolate the problem into independent conditions. For each independent condition you calculate the combinations of partitions, then you multiply them together.
I have walked through this example manually, but you can program the same procedure.