I implementing Family program in prolog i have a problem to implement some rules.
first of all i implement this rule:
number_of_children_couple(_list):-
findall(children(_f,_m,_n),children(_f,_m,_n),_list).
return list:
19 ?- number_of_children_couple(_list).
_list = [children(mordechai, miriam, 1), children(salax, naima, 1), children(eli, bella, 2), children(..., ..., ...)|...].
my problem is how to implement :
number_of_children_person(_list28,list_person):-
first argument:
_list28 = _list //the above list ,that return from the rule
and second argument is :
list_person = [children(mordechai, 1), children(salax, 1),children(eli, 2),children(..., ...)|...]
and i also use with:
%_num is number of children for couple
children(_father,_mother,_num):-
couple(_mother,_father),
findall(_child,parents(_mother,_father,_child),_children),
length1(_children,_num).
%_num is number of children for _person
children(_person,_num):-
gender(_person,_),
findall(_child,parent(_person,_child),_list),
length1(_list,_num).
if what you want is to drop an argument from children/3 structure, can be done in a number of ways, the simpler being
number_of_children_person([],[]).
number_of_children_person([children(A,_,C)|R],[children(A,C)|T]) :-
number_of_children_person(R,T).
or more succintly
number_of_children_person(L3,L2) :-
findall(children(A,C), member(children(A,B,C),L3), L2).
Related
I have two lists and I want to return a result in the following way:
the result should contain elements that are in list one and list two
output should be same order as per first list
Input :
val first = listOf(1, 2, 3, 4, 5,7,9,15,11)
val second = listOf(2, 15 , 4,3, 11)
Output:
val output = listOf(2,3,4,15,11)
Please help me to learn how to get common values in both lists in order of list first in Kotlin.
You can do
val output = first.filter { second.contains(it) }
What you are looking for is the intersection of the two lists:
val output = first.intersect(second)
As pointed out by #Ivo the result is a Set which can be turned into a list with output.toList(). However, since the result is a set, it contains no duplicates, e.g. if first is listOf(1,2,3,1,2,3) and second is listOf(2,4,2,4), the result will be equal to setOf(2).
If this is not acceptable, the solution of #Ivo should be used instead.
I have a list of tuples structured as per the below (data is example only):
[('aaa', 10), ('bbb', 10), ('ccc', 12), ('ddd', 12), ('eee', 14)]
I need to search the second item in each of the tuples (the number) to see if it exists in the list (eg search for 12 = found, search for 5 = not found.
Currently I am using the below, which works but may not be the best way in Python:
not_there = True
for a in final_set:
if final_set[1] == episode_id:
not_there = False
break
What is the best / most efficient way in Python to do this?
Maybe you can try something like this :
test = [('aaa', 10), ('bbb', 10), ('ccc', 12), ('ddd', 12), ('eee', 14)]
number = 10
for i in test:
if number in i:
print("Number {} found.".format(number))
else:
print("Number {} not found".format(number))
That should work regardless you're searching element 1 in the tuple (the 'aaa') or element 2 (the number).
Hope this helps.
What about this:
is_there = (len([item for item in final_set if item[1] == episode_id]) > 0)
Basically, [item for item in final_set if item[1] == episode_id] is a list comprehension expression which creates a list of the items in final_set such that item[1] == episode_id.
Then, you can check the length of the resulting list: if it is greater than 0, than something has been found.
I have a couple List<string>s, with the format like this:
List 1 List 2 List 3
1 A One
2 B Two
3 C Three
4 D Four
5 E Five
So in code form, it's like:
List<string> list1 = {"1","2","3","4","5"};
List<string> list2 = {"A","B","C","D","E"};
List<string> list3 = {"One","Two","Three","Four","Five"};
My questions are:
How do I transfom those three lists to a CSV format?
list1,list2,list3
1,A,one
2,b,two
3,c,three
4,d,four
5,e,five
Should I append , to the end of each index or make the delimeter its own index within the multidimensional list?
If performance is your main concern, I would use an existing csv library for your language, as it's probably been pretty well optimized.
If that's too much overhead, and you just want a simple function, I use the same concept in some of my code. I use the join/implode function of a language to create a list of comma separated strings, then join that list with \n.
I'm used to doing this in a dynamic language, but you can see the concept in the following pseudocode example:
header = {"List1", "List2", "List3"}
list1 = {"1","2","3","4","5"};
list2 = {"A","B","C","D","E"};
list3 = {"One","Two","Three","Four","Five"};
values = {header, list1, list2, list3};
for index in values
values[index] = values[index].join(",");
values = values.join("\n");
I have a Python list like the following:
['IKW', 'IQW', 'IWK', 'IWQ', 'KIW', 'KLW', 'KWI', 'KWL', 'LKW', 'LQW', 'LWK', 'LWQ', 'QIW', 'QLW', 'QWI', 'QWL', 'WIK', 'WIQ', 'WKI', 'WKL', 'WLK', 'WLQ', 'WQI', 'WQL']
If we pick, say the second element IQW, we see that the list has duplicates of this item HOWEVER its not noticeable right away. This is because it is cyclic. I mean the following are equivalent.
IQW, QWI, WIQ
Also it could be backwards which is also a duplicate so I want it removed. So now the list of duplicates are (the reverse of each of one these)
IQW, QWI, WIQ , WQI, IWQ, QIW
So essentially I would like IQW to be the only one left.
Bonus points, if the one that is remaining in the list is sorted alphabetically.
The way I did was to sort the entire list by alphabetical order:
`IQW`, `QWI`, `WIQ` , `WQI`, `IWQ`, `QIW` ->
`IQW`, `IQW`, `IQW`, `IQW`, `IQW` `IQW`
and then remove the duplicates.
However this also removes combinations say i have ABCD and CDAB. These are not the same because the ends only meet once. But my method will sort them to ABCD and ABCD and remove one.
My code:
print cur_list
sortedlist = list()
for i in range(len(cur_list)):
sortedlist.append(''.join(map(str, sorted(cur_list[i]))))
sortedlist = set(sortedlist)
L = ['IKW', 'IQW', 'IWK', 'IWQ', 'KIW', 'KLW', 'KWI', 'KWL', 'LKW', 'LQW', 'LWK', 'LWQ', 'QIW', 'QLW', 'QWI', 'QWL', 'WIK', 'WIQ', 'WKI', 'WKL', 'WLK', 'WLQ', 'WQI', 'WQL']
seen = set()
res = []
for item in L:
c = item.index(min(item))
item = item[c:] + item[:c]
if item not in seen:
seen.add(item)
seen.add(item[0]+item[-1:0:-1])
res.append(item)
print res
output:
['IKW', 'IQW', 'KLW', 'LQW']
Here is the solution I coded: If anyone has a better algo, I will accept that as answer:
mylist = list()
for item in copy_of_cur:
linear_peptide = item+item
mylist = filter(lambda x: len(x) == 3 , subpeptides_linear(linear_peptide))
for subitem in mylist:
if subitem != item:
if subitem in cur_list:
cur_list.remove(subitem)
I'm trying to do two things. (1) display each element of an existing list, and (2) search a list to display all names that contain that element.
Here are some facts:
classes(hannes, [cs490, cs499, cs413]). % name has this list of classes
classes(tony, [ma330, ma211, ma250]).
classes(nicholas, [cs424, cs570, ma330]).
classes(pj, [ma211, ma250, ma285, cs424]).
classes(inga, [cs285, cs307, cs309]).
classes(christine, [ma285, ma211, ma330]).
classes(lisa, [cs424, cs413, cs490]).
classes(marty, [cs570, cs424]).
And, here is my rule so far:
taking(N,C) :- % student Name N is taking class C
classes(N,Cs),
[C|T] = Cs.
At the moment, I know this only takes the head of the list and displays it. I need to display each item of the list (one line at a time, if easy enough to do, but not important). But, I also need to be able to do it in reverse. If 1 course is queried (ma330), I want it to display all students that have that particular course.
Query example 1:
?- taking(nicholas, Classes).
Classes = [cs424, cs570, ma330] ;
OR
?- taking(nicholas, Classes).
Classes = cs424 ;
Classes = cs570 ;
Classes = ma330 ;
Query example 2:
?- taking(Names, ma330).
Names = tony ;
Names = nicholas ;
Names = christine ;
I'm going to keep searching for a resolution, but if anyone can help, it would be appreciated.
Thank you!!!
Think of that : C is member of Classes.
EDIT OK try this code :
taking(N,C) :- % student Name N is taking class C
classes(N,Cs),
member(C, Cs).