I have a feed that I want to split based on the number of the item. For example, split it so it outputs items 5 - 10 or 4 - 12. I'm new to Yahoo Pipes, so is this possible, and if so, how?
Use the tail and revese operators
exemple to extract 5- 10 items
you have a feed :
1 / 2 / ...
reverse
=> ... / 2 / 1
tail 10
=> 10 / ... / 1
reverse
=> 1 / ... /10
tail 6
-> 5 / 6 / 7 / 8 / 9 / 10
Related
Does inserting order affects the structure of binary heap? I mean, is it possible to get a little different parent-children relations when inserting the same elements in different orders,
for example:
20 6 3 5 7 8 16 10 (inserting order #1) and 6 3 20 10 16 3 7 5 (inserting order #2)
or the final result should always be the same?
A heap can store a given collection of values in different ways. For instance, if the heap happens to be a perfect binary tree, then you can swap any subtree with its sibling subtree without violating the heap property.
For example, if the data collection has the values 1, 2 and 3, there are 2 possible heaps that can represent that data set:
1 1
/ \ / \
2 3 3 2
The first will be the result when 2 is inserted before 3, and the second heap will be the result when 2 is inserted after 3.
If we look at an input with four values (e.g. 1, 2, 3 and 4), we can represent that in four heaps:
1 1 1 1
/ \ / \ / \ / \
2 3 2 4 3 2 4 2
/ / / /
4 3 4 3
Again, the order of insertion will determine which of those four heaps will be the end result.
If you imagine the sequences 1 1 1 1 2 2 and 1 1 2 1 1 2, you should end up with different heaps.
1 versus 1
1 1 1 2
1 2 2 1 1 2
I have got a sequence 1 2 3 4 5 6 ... n. Now, I am given a sequence of n deletions - each deletion is a number which I want to delete. I need to respond to each deletion with two numbers - of a left and right neighbour of deleted number (-1 if any doesn't exists).
E.g. I delete 2 - I respond 1 3, then I delete 3 I respond 1 4 , I delete 6 I respond 5 -1 etc.
I want to do it fast - linear of linear-logarithmic time complexity.
What data structure should I use? I guess the key to the solution is the fact that the sequence is sorted.
A doubly-linked list will do fine.
We will store the links in two arrays, prev and next, to allow O(1) access for deletions.
First, for every element and two sentinels at the ends, link it to the previous and next integers:
init ():
for cur := 0, 1, 2, ..., n, n+1:
prev[cur] := cur-1
next[cur] := cur+1
When you delete an element cur, update the links in O(1) like this:
remove (cur):
print (num (prev[cur]), " ", num (next[cur]), newline)
prev[next[cur]] := prev[cur]
next[prev[cur]] := next[cur]
Here, the num wrapper is inserted to print -1 for the sentinels:
num (cur):
if (cur == 0) or (cur == n+1):
return -1
else:
return cur
Here's how it works:
prev next
n = 6 prev/ print 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7
/next ------------------- -------------------
init () -1 0 1 2 3 4 5 6 1 2 3 4 5 6 7 8
remove (2) 1 3 1 3 -1 0 1 3 4 5 6 1 3 4 5 6 7 8
remove (3) 1 4 1 4 -1 0 1 4 5 6 1 4 5 6 7 8
remove (6) 5 7 5 -1 -1 0 1 4 5 1 4 5 7 8
remove (1) 0 4 -1 4 -1 0 4 5 4 5 7 8
remove (5) 4 7 4 -1 -1 0 4 4 7 8
remove (4) 0 7 -1 -1 -1 0 7 8
Above, the portions not used anymore are blanked out for clarity.
The respective elements of the arrays still store the values printed above them, but we no longer access them.
As Jim Mischel rightly noted (thanks!), storing the list in two arrays instead of dynamically allocating the storage is crucial to make this O(1) per deletion.
You can use a binary search tree. Deleting from it is logarithmic. If you want to remove n elements and the number of total elements is m, then the complexity of removing n elements from it will be
nlogm
I have records like these:
1 4 6 4 2 4 8
2 3 5 4 6 7 1
5 4 6 4 3 8 4
1 4 6 4 5 7 1
5 7 3 3 3 6 3
6 7 3 3 4 8 4
I want to sort them on columns 2,3,4, and 6 and keep just one of those identical in column 2,3,4 and having the biggest number in column 6 such as:
1 4 6 4 5 7 1
2 3 5 4 6 7 1
5 4 6 4 3 8 4
5 7 3 3 3 6 3
6 7 3 3 4 8 4
I have tried all kinds of combinations between sort and uniq but everything fails because uniq cannot be applied onto a specific column. The only thing I came up with is to change the order of the columns as to first sort as above then move records 2,3,and 4 to the end and then run uniq with -w as to focus only on the last 3 records. This seems quite inefficient to me.
Thanks for help!
You can achieve this with two passes of sort(assuming in the first place I understand your requirement correctly, seeing that the desired data snippet posted above does not match your description of it) . The first pass sorts by field 2 through 4 ascending and field 6 descending, the second pass sorts on fields 2 through 4 only but passing in the "stable sort" and unique flags in addition to pick out those rows for each combination of fields 2-4 that have the highest value from field 6
sort -k2,4n -k6,6nr file.txt | sort -k2,4n -s -u
2 3 5 4 6 7 1
5 4 6 4 3 8 4
6 7 3 3 4 8 4
I want to parse a text file, where
I get numbers that are between parenthesis like
this:
1 2 3 (4 - 7) 8 9
1 3 8 (7 - 8) 2 1
1 2 (8 - 10) 3 2
should return an array for each:
array1:
4
7
8
array2:
7
8
10
I am thinking of using split for each line, like line.split("("), but that doesn't quite doing the trick.. I was wondering if there is something more sophisticated for the job.
Any help appreciated,
Ted
data = <<EOS
1 2 3 (4 - 7) 8 9
1 3 8 (7 - 8) 2 1
1 2 (8 - 10) 3 2
EOS
lines = data.split("\n")
def get_inner(lines)
lines.map { |line| line.partition("(")[2].partition(")")[0].split(" - ")}
end
a1, a2 = *[get_inner(lines).map {|a| a.first },get_inner(lines).map {|a| a.last }]
puts a1
puts a2
# =>
4
7
8
7
8
10
I would look into using things like Substring / IndexOf as well as split.
You could also try a regular expression to find numbers seperated by spaces in between the ( ) but regular expressions can be a bit of a pain.
Hmm just found this
http://www.rubular.com/ I got the expression I needed
((\d+)-(\d+))
I have array say "a"
a =
1 4 5
6 7 2
if i use function
b=sort(a)
gives ans
b =
1 4 2
6 7 5
but i want ans like
b =
5 1 4
2 6 7
mean 2nd row should be sorted but elements of ist row should remain unchanged and should be correspondent to row 2nd.
sortrows(a',2)'
Pulling this apart:
a = 1 4 5
6 7 2
a' = 1 6
4 7
5 2
sortrows(a',2) = 5 2
1 6
4 7
sortrows(a',2)' = 5 1 4
2 6 7
The key here is sortrows sorts by a specified row, all the others follow its order.
You can use the SORT function on just the second row, then use the index output to sort the whole array:
[junk,sortIndex] = sort(a(2,:));
b = a(:,sortIndex);
How about
a = [1 4 5; 6 7 2]
a =
1 4 5
6 7 2
>> [s,idx] = sort(a(2,:))
s =
2 6 7
idx =
3 1 2
>> b = a(:,idx)
b =
5 1 4
2 6 7
in other words, you use the second argument of sort to get the sort order you want, and then you apply it to the whole thing.