I'm trying to loop over two lists, one being a premade list and one being a list literal. Is something like this possible?
Pseudocode example:
list(list1 APPEND 0 1 2 3 4)
foreach(item IN LISTS ${list1} 5 6 7 8 9)
message(${item} ${#other variable})
endforeach(item)
# prints out
0 5
1 6
... etc
For iterate over several lists at the same time, you may use foreach loop over their indicies. Then, in the loop body access the lists's elements by that index:
# Setup content of the lists somehow
set(list1 0 1 2 3 4)
set(list2 5 6 7 8 9)
list(LENGTH list1 n_elems) # Total number of the elements in the every list
math(EXPR last_index "${n_elems}-1") # The last index in the every list
# Now iterate over indicies
foreach(i RANGE ${last_index})
list(GET list1 ${i} elem1) # Element in the first list
list(GET list2 ${i} elem2) # Corresponded element in the second list
# Do something with elements
message("${elem1} ${elem2}")
endforeach()
Related
I have a DataFrame, in which I want to merge certain rows to a single one. It has the following structure (values repeat)
Index Value
1 date:xxxx
2 user:xxxx
3 time:xxxx
4 description:xxx1
5 xxx2
6 xxx3
7 billed:xxxx
...
Now the problem is, that the columns 5 & 6 still belong to the description and were separated just wrong (whole string separated by ","). I want to merge the "description" row (4) with the values afterwards (5,6). In my DF, there can be 1-5 additional entries which have to be merged with the description row, but the structure allows me to work with startswith, because no matter how many rows have to be merged, the end point is always the row which starts with "billed". Due to me being very new to python, I haven´t got any code written for this problem yet.
My thought is the following (if it is even possible):
Look for a row which starts with "description" → Merge all the rows afterwards till reaching the row which starts with "billed", then stop (obviosly we keep the "billed" row) → Do the same to each row starting with "description"
New DF should look like:
Index Value
1 date:xxxx
2 user:xxxx
3 time:xxxx
4 description:xxx1, xxx2, xxx3
5 billed:xxxx
...
df = pd.DataFrame.from_dict({'Value': ('date:xxxx', 'user:xxxx', 'time:xxxx', 'description:xxx', 'xxx2', 'xxx3', 'billed:xxxx')})
records = []
description = description_val = None
for rec in df.to_dict('records'): # type: dict
# if previous description and record startswith previous description value
if description and rec['Value'].startswith(description_val):
description['Value'] += ', ' + rec['Value'] # add record Value into previous description
continue
# record with new description...
if rec['Value'].startswith('description:'):
description = rec
_, description_val = rec['Value'].split(':')
elif rec['Value'].startswith('billed:'):
# billed record - remove description value
description = description_val = None
records.append(rec)
print(pd.DataFrame(records))
# Value
# 0 date:xxxx
# 1 user:xxxx
# 2 time:xxxx
# 3 description:xxx, xxx2, xxx3
# 4 billed:xxxx
I'd like to find the contiguous sequences of equal elements (e.g. of length 2) in a list
my #s = <1 1 0 2 0 2 1 2 2 2 4 4 3 3>;
say grep {$^a eq $^b}, #s;
# ==> ((1 1) (2 2) (4 4) (3 3))
This code looks ok but when one more 2 is added after the sequence of 2 2 2 or when one 2 is removed from it, it says Too few positionals passed; expected 2 arguments but got 1 How to fix it? Please note that I'm trying to find them without using for loop, i.e. I'm trying to find them using a functional code as much as possible.
Optional: In the bold printed section:
<1 1 0 2 0 2 1 2 2 2 4 4 3 3>
multiple sequences of 2 2 are seen. How to print them the number of times they are seen? Like:
((1 1) (2 2) (2 2) (4 4) (3 3))
There are an even number of elements in your input:
say elems <1 1 0 2 0 2 1 2 2 2 4 4 3 3>; # 14
Your grep block consumes two elements each time:
{$^a eq $^b}
So if you add or remove an element you'll get the error you're getting when the block is run on the single element left over at the end.
There are many ways to solve your problem.
But you also asked about the option of allowing for overlapping so, for example, you get two (2 2) sub-lists when the sequence 2 2 2 is encountered. And, in a similar vein, you presumably want to see two matches, not zero, with input like:
<1 2 2 3 3 4>
So I'll focus on solutions that deal with those issues too.
Despite the narrowing of solution space to deal with the extra issues, there are still many ways to express solutions functionally.
One way that just appends a bit more code to the end of yours:
my #s = <1 1 0 2 0 2 1 2 2 2 4 4 3 3>;
say grep {$^a eq $^b}, #s .rotor( 2 => -1 ) .flat
The .rotor method converts a list into a list of sub-lists, each of the same length. For example, say <1 2 3 4> .rotor: 2 displays ((1 2) (3 4)). If the length argument is a pair, then the key is the length and the value is an offset for starting the next pair. If the offset is negative you get sub-list overlap. Thus say <1 2 3 4> .rotor: 2 => -1 displays ((1 2) (2 3) (3 4)).
The .flat method "flattens" its invocant. For example, say ((1,2),(2,3),(3,4)) .flat displays (1 2 2 3 3 4).
A perhaps more readable way to write the above solution would be to omit the flat and use .[0] and .[1] to index into the sub-lists returned by rotor:
say #s .rotor( 2 => -1 ) .grep: { .[0] eq .[1] }
See also Elizabeth Mattijsen's comment for another variation that generalizes for any sub-list size.
If you needed a more general coding pattern you might write something like:
say #s .pairs .map: { .value xx 2 if .key < #s - 1 and [eq] #s[.key,.key+1] }
The .pairs method on a list returns a list of pairs, each pair corresponding to each of the elements in its invocant list. The .key of each pair is the index of the element in the invocant list; the .value is the value of the element.
.value xx 2 could have been written .value, .value. (See xx.)
#s - 1 is the number of elements in #s minus 1.
The [eq] in [eq] list is a reduction.
If you need text pattern matching to decide what constitutes contiguous equal elements you might convert the input list into a string, match against that using one of the match adverbs that generate a list of matches, then map from the resulting list of matches to your desired result. To match with overlaps (eg 2 2 2 results in ((2 2) (2 2)) use :ov:
say #s .Str .match( / (.) ' ' $0 /, :ov ) .map: { .[0].Str xx 2 }
TIMTOWDI!
Here's an iterative approach using gather/take.
say gather for <1 1 0 2 0 2 1 2 2 2 4 4 3 3> {
state $last = '';
take ($last, $_) if $last == $_;
$last = $_;
};
# ((1 1) (2 2) (2 2) (4 4) (3 3))
> my #numbers = 1, 3, 5;
> 1 ~~ /#numbers/; #
｢1｣
is the same as:
> 1 ~~ /1 | 3 | 5/
｢1｣
but when the element is a Range object, it fails to match:
> my #ranges = 1..3.item, 4..6.item;
[1..3 4..6]
> 1 ~~ /#ranges/
Nil
> 1 ~~ /|#ranges/
Nil
> 1 ~~ /||#ranges/
When the regex engine sees /#numbers/ it treats that like an alternation of the array elements, so your first two examples are equivalent.
There just is no such automatism for Ranges I believe.
Edit: Never mind below, I totally misread the question at first.
> my #ranges = 1..3, 4..6;
[1..3 4..6]
> 1 ~~ #ranges[0];
True
> 2 ~~ #ranges[1];
False
> 4 ~~ #ranges[1];
True
> #ranges.first( 5 ~~ * )
4..6
See? #ranges is a array of, well, ranges (your call to item does nothing here). Theoretically this would hold true if the smartmatch operator were smarter.
> 1..3 ~~ #ranges;
False
Flattening also doesn't help, because a flat list of ranges is still a list of ranges.
Flattening the ranges themselves is possible, but that simply turns them into Arrays
> my #ranges2 = |(1..3), |(4..6)
[1 2 3 4 5 6]
Why does single number fails to match Range object in array?
Per the doc:
The interpolation rules for individual elements [of an array] are the same as for scalars
And per the same doc section the rule for a scalar (that is not a regex) is:
interpolate the stringified value
A range object such as 1..3 stringifies to 1 2 3:
my $range = 1..3;
put $range; # 1 2 3
put so '1' ~~ / $range /; # False
put so '1 2 3' ~~ / $range /; # True
So, as Holli suggests, perhaps instead:
my #ranges = flat 1..3, 4..6;
say #ranges; # [1 2 3 4 5 6]
say 1 ~~ /#ranges/; # ｢1｣
Or is there some reason you don't want that? (See also Scimon's comment on Holli's answer.)
my #numbers = <4 8 15 16 23 42>;
this works:
.say for #numbers[0..2]
# 4
# 8
# 15
but this doesn't:
my $range = 0..2;
.say for #numbers[$range];
# 16
the subscript seems to be interpreting $range as the number of elements in the range (3). what gives?
Working as intended. Flatten the range object into a list with #numbers[|$range] or use binding on Range objects to hand them around. https://docs.perl6.org will be updated shortly.
On Fri Jul 22 15:34:02 2016, gfldex wrote:
> my #numbers = <4 8 15 16 23 42>; my $range = 0..2; .say for
> #numbers[$range];
> # OUTPUT«16»
> # expected:
> # OUTPUT«4815»
>
This is correct, and part of the "Scalar container implies item" rule.
Changing it would break things like the second evaluation here:
> my #x = 1..10; my #y := 1..3; #x[#y]
(2 3 4)
> #x[item #y]
4
Noting that since a range can bind to #y in a signature, then Range being a
special case would make an expression like #x[$(#arr-param)]
unpredictable in its semantics.
> # also binding to $range provides the expected result
> my #numbers = <4 8 15 16 23 42>; my $range := 0..2; .say for
> #numbers[$range];
> # OUTPUT«4815»
> y
This is also expected, since with binding there is no Scalar container to
enforce treatment as an item.
So, all here is working as designed.
A symbol bound to a Scalar container yields one thing
Options for getting what you want include:
Prefix with # to get a plural view of the single thing: numbers[#$range]; OR
declare the range variable differently so it works directly
For the latter option, consider the following:
# Bind the symbol `numbers` to the value 1..10:
my \numbers = [0,1,2,3,4,5,6,7,8,9,10];
# Bind the symbol `rangeA` to the value 1..10:
my \rangeA := 1..10;
# Bind the symbol `rangeB` to the value 1..10:
my \rangeB = 1..10;
# Bind the symbol `$rangeC` to the value 1..10:
my $rangeC := 1..10;
# Bind the symbol `$rangeD` to a Scalar container
# and then store the value 1..10 in it:`
my $rangeD = 1..10;
# Bind the symbol `#rangeE` to the value 1..10:
my #rangeE := 1..10;
# Bind the symbol `#rangeF` to an Array container and then
# store 1 thru 10 in the Scalar containers 1 thru 10 inside the Array
my #rangeF = 1..10;
say numbers[rangeA]; # (1 2 3 4 5 6 7 8 9 10)
say numbers[rangeB]; # (1 2 3 4 5 6 7 8 9 10)
say numbers[$rangeC]; # (1 2 3 4 5 6 7 8 9 10)
say numbers[$rangeD]; # 10
say numbers[#rangeE]; # (1 2 3 4 5 6 7 8 9 10)
say numbers[#rangeF]; # (1 2 3 4 5 6 7 8 9 10)
A symbol that's bound to a Scalar container ($rangeD) always yields a single value. In a [...] subscript that single value must be a number. And a range, treated as a single number, yields the length of that range.
I'm trying to create 20 unique cards with numbers, but I struggle a bit.. So basically I need to create 20 unique matrices 3x3 having numbers 1-10 in first column, numbers 11-20 in the second column and 21-30 in the third column.. Any ideas? I'd prefer to have it done in r, especially as I don't know Visual Basic. In excel I know how to generate the cards, but not sure how to ensure they are unique..
It seems to be quite precise and straightforward to me. Anyway, i needed to create 20 matrices that would look like :
[,1] [,2] [,3]
[1,] 5 17 23
[2,] 8 18 22
[3,] 3 16 24
Each of the matrices should be unique and each of the columns should consist of three unique numbers ( the 1st column - numbers 1-10, the 2nd column 11-20, the 3rd column - 21-30).
Generating random numbers is easy, though how to make sure that generated cards are unique?Please have a look at the post that i voted for as an answer - as it gives you thorough explanation how to achieve it.
(N.B. : I misread "rows" instead of "columns", so the following code and explanation will deal with matrices with random numbers 1-10 on 1st row, 11-20 on 2nd row etc., instead of columns, but it's exactly the same just transposed)
This code should guarantee uniqueness and good randomness :
library(gtools)
# helper function
getKthPermWithRep <- function(k,n,r){
k <- k - 1
if(n^r< k){
stop('k is greater than possibile permutations')
}
v <- rep.int(0,r)
index <- length(v)
while ( k != 0 )
{
remainder<- k %% n
k <- k %/% n
v[index] <- remainder
index <- index - 1
}
return(v+1)
}
# get all possible permutations of 10 elements taken 3 at a time
# (singlerowperms = 720)
allperms <- permutations(10,3)
singlerowperms <- nrow(allperms)
# get 20 random and unique bingo cards
cards <- lapply(sample.int(singlerowperms^3,20),FUN=function(k){
perm2use <- getKthPermWithRep(k,singlerowperms,3)
m <- allperms[perm2use,]
m[2,] <- m[2,] + 10
m[3,] <- m[3,] + 20
return(m)
# if you want transpose the result just do:
# return(t(m))
})
Explanation
(disclaimer tl;dr)
To guarantee both randomness and uniqueness, one safe approach is generating all the possibile bingo cards and then choose randomly among them without replacements.
To generate all the possible cards, we should :
generate all the possibilities for each row of 3 elements
get the cartesian product of them
Step (1) can be easily obtained using function permutations of package gtools (see the object allPerms in the code). Note that we just need the permutations for the first row (i.e. 3 elements taken from 1-10) since the permutations of the other rows can be easily obtained from the first by adding 10 and 20 respectively.
Step (2) is also easy to get in R, but let's first consider how many possibilities will be generated. Step (1) returned 720 cases for each row, so, in the end we will have 720*720*720 = 720^3 = 373248000 possible bingo cards!
Generate all of them is not practical since the occupied memory would be huge, thus we need to find a way to get 20 random elements in this big range of possibilities without actually keeping them in memory.
The solution comes from the function getKthPermWithRep, which, given an index k, it returns the k-th permutation with repetition of r elements taken from 1:n (note that in this case permutation with repetition corresponds to the cartesian product).
e.g.
# all permutations with repetition of 2 elements in 1:3 are
permutations(n = 3, r = 2,repeats.allowed = TRUE)
# [,1] [,2]
# [1,] 1 1
# [2,] 1 2
# [3,] 1 3
# [4,] 2 1
# [5,] 2 2
# [6,] 2 3
# [7,] 3 1
# [8,] 3 2
# [9,] 3 3
# using the getKthPermWithRep you can get directly the k-th permutation you want :
getKthPermWithRep(k=4,n=3,r=2)
# [1] 2 1
getKthPermWithRep(k=8,n=3,r=2)
# [1] 3 2
Hence now we just choose 20 random indexes in the range 1:720^3 (using sample.int function), then for each of them we get the corresponding permutation of 3 numbers taken from 1:720 using function getKthPermWithRep.
Finally these triplets of numbers, can be converted to actual card rows by using them as indexes to subset allPerms and get our final matrix (after, of course, adding +10 and +20 to the 2nd and 3rd row).
Bonus
Explanation of getKthPermWithRep
If you look at the example above (permutations with repetition of 2 elements in 1:3), and subtract 1 to all number of the results you get this :
> permutations(n = 3, r = 2,repeats.allowed = T) - 1
[,1] [,2]
[1,] 0 0
[2,] 0 1
[3,] 0 2
[4,] 1 0
[5,] 1 1
[6,] 1 2
[7,] 2 0
[8,] 2 1
[9,] 2 2
If you consider each number of each row as a number digit, you can notice that those rows (00, 01, 02...) are all the numbers from 0 to 8, represented in base 3 (yes, 3 as n). So, when you ask the k-th permutation with repetition of r elements in 1:n, you are also asking to translate k-1 into base n and return the digits increased by 1.
Therefore, given the algorithm to change any number from base 10 to base n :
changeBase <- function(num,base){
v <- NULL
while ( num != 0 )
{
remainder = num %% base # assume K > 1
num = num %/% base # integer division
v <- c(remainder,v)
}
if(is.null(v)){
return(0)
}
return(v)
}
you can easily obtain getKthPermWithRep function.
One 3x3 matrix with the desired value range can be generated with the following code:
mat <- matrix(c(sample(1:10,3), sample(11:20,3), sample(21:30, 3)), nrow=3)
Furthermore, you can use a for loop to generate a list of 20 unique matrices as follows:
for (i in 1:20) {
mat[[i]] <- list(matrix(c(sample(1:10,3), sample(11:20,3), sample(21:30,3)), nrow=3))
print(mat[[i]])
}
Well OK I may fall on my face here but I propose a checksum (using Excel).
This is a unique signature for each bingo card which will remain invariate if the order of numbers within any column is changed without changing the actual numbers. The formula is
=SUM(10^MOD(A2:A4,10)+2*10^MOD(B2:B4,10)+4*10^MOD(C2:C4,10))
where the bingo numbers for the first card are in A2:C4.
The idea is to generate a 10-digit number for each column, then multiply each by a constant and add them to get the signature.
So here I have generated two random bingo cards using a standard formula from here plus two which are deliberately made to be just permutations of each other.
Then I check if any of the signatures are duplicates using the formula
=MAX(COUNTIF(D5:D20,D5:D20))
which shouldn't given an answer more than 1.
In the unlikely event that there were duplicates, then you would just press F9 and generate some new cards.
All formulae are array formulae and must be entered with CtrlShiftEnter
Here is an inelegant way to do this. Generate all possible combinations and then sample without replacement. These are permutations, combinations: order does matter in bingo
library(dplyr)
library(tidyr)
library(magrittr)
generate_samples = function(n) {
first = data_frame(first = (n-9):n)
first %>%
merge(first %>% rename(second = first)) %>%
merge(first %>% rename(third = first)) %>%
sample_n(20)
}
suffix = function(df, suffix)
df %>%
setNames(names(.) %>%
paste0(suffix))
generate_samples(10) %>% suffix(10) %>%
bind_cols(generate_samples(20) %>% suffix(20)) %>%
bind_cols(generate_samples(30) %>% suffix(30)) %>%
rowwise %>%
do(matrix = t(.) %>% matrix(3)) %>%
use_series(matrix)