Question: Let's say we have a List<E> in Groovy. How can we convert it into a Tuple<E>?
As beginner in Groovy, I could not find the Groovy equivalent of the following Python code:
a = [1, 2, "three"]
b = tuple(a)
print(b)
# (1, 2, 'three')
print(type(b))
# <class 'tuple'>
Two ways, which seemed plausible, did not return exactly the same results:
a = [1, 2, "three"]
// First way, cast as
b = a as Tuple
print(b)
// [[1, 2, three]]
// Second way, asImmutable()
b = a.asImmutable()
print(b)
// [1, 2, three]
b.getClass()
// class java.util.Collections$UnmodifiableRandomAccessList
So, as Tuple returned a tuple containing the original list as a single element, whereas .asImmutable() returned an UnmodifiableRandomAccessList, which is almost like a tuple but not a tuple.
Is it not possible to create a tuple from a list in Groovy? If it is, how can it be done?
Related
In Dart programming language, sometimes I see functions returning a list in pair of brackets with triple dots [...list]. For example:
class IntList {
List<int> _integerList = [1, 2, 3, 4, 5];
List<int> get integerList {
return [..._integerList];
}
}
So what is the difference between return integerList; and the above return statement?
Any help is really appreciated. Thank you.
For this particular case there is no difference. ... is the spread operator. This allows you to insert multiple elements into a collection.
For example:
var list = [1, 2, 3];
var list2 = [0, ...list];
print(list2)
Output:
[0, 1, 2, 3]
So doing return [..._integerList]; is the exact same as return integerList;, other than that it creates a new list.
var list = [1, 2, 3];
print(list.hashCode);
print([...list].hashCode);
This code shows that they are different List objects when the spread operator is used because of the output of different hash codes.
e.g. [a,a,c,e] and [a,b,c,d,e]. This is what I mean by same order. It shouldn't matter if one of the lists is only partial. To be clear, I am uninterested in sorting anything.
Edit: Apparently I should mention that I mean lists with some different values. e.g. [a,a,c,e] and [a,b,c,d,e] are in the same order. Simply getting rid of duplicates and cutting the bigger list won't work.
We can use itertools.groupby to squash identical consecutive items into a single item of that value. We can then zip together the squashed values with the other list to see if they are the same
from itertools import izip, groupby
def squash(iterable):
return (k for k, _ in groupby(iterable))
def same_order(iterable, reference):
return all(x == y for x, y in zip(squash(iterable), reference))
print(same_order((1, 2, 2, 3, 4), (1, 2, 3, 4, 5)))
# True
print(same_order((1, 2, 2, 1, 3, 4), (1, 2, 3, 4, 5)))
# False
You can first take the unique set of b, then truncate to the size of a (or vice versa)
a = [1,2,3,4]
b = [1,1,2,2,3,4,4,5]
a == list(set(b))[:len(a)]
You can achieve this by first getting the unique elements from partial list a (in the order of a), then get those elements from full list b (in the order of b). Then you can compare the two and see if they are identical.
>>> a = [1, 1, 4, 3]
>>> a = sorted(set(a), key=a.index) # Remove duplicate elements, but keep the original order
>>> a
[1, 4, 3]
>>> b = [1, 2, 4, 3, 5]
>>> b = [x for x in sorted(set(b), key=b.index) if x in a] # Get unique elements of a from b in order of b, and filter them out to get only elements present in a
>>> a == b
True
>>> a = [4, 3, 2, 1]
>>> a = sorted(set(a), key=a.index) # Remove duplicate elements, but keep the original order
>>> b = [1, 2, 4, 3, 5]
>>> b = [x for x in sorted(set(b), key=b.index) if x in a] # Get unique elements of a from b in order of b, and filter them out to get only elements present in a
>>> a == b
False
I want remove list b from list a without changing id of list a I tried below method:
for i in b:
a.remove(i)
Is there any better ways?
Something like the below?
a = [1, 2, 3]
b = [4, 5, 3]
for i in b:
try:
a.remove(i)
except ValueError:
pass
print a
Alternatively:
a = [i for i in a if i not in b]
print a
If I have an input like this (1, 2, 3, 4, 5, 6)
The output has to be ... [[1, 2], [3, 4], [5, 6]].
I know how to deal with if it's one element but not two.
x=[]
for number in numbers:
x.append([number])
I'll appreciate your any help!
Something like this would work:
out = []
lst = (1,2,3,4,5,6,7,8,9,10)
for x in range(len(lst)):
if x % 2 == 0:
out.append([lst[x], lst[x+1]])
else:
continue
To use this, just set lst equal to whatever list of numbers you want. The final product is stored in out.
There is a shorter way of doing what you want:
result = []
L = (1,2,3,4,5,6,7,8,9,10)
result = [[L[i], L[i + 1]] for i in range(0, len(L) - 1, 2)]
print(result)
You can use something like this. This solution also works for list of odd length
def func(lst):
res = []
# Go through every 2nd value | 0, 2, 4, ...
for i in range(0, len(lst), 2):
# Append a slice of the list, + 2 to include the next value
res.append(lst[i : i + 2])
return res
# Output
>>> lst = [1, 2, 3, 4, 5, 6]
>>> func(lst)
[[1, 2], [3, 4], [5, 6]]
>>> lst2 = [1, 2, 3, 4, 5, 6, 7]
>>> func(lst2)
[[1, 2], [3, 4], [5, 6], [7]]
List comprehension solution
def func(lst):
return [lst[i:i+2] for i in range(0, len(lst), 2)]
Slicing is better in this case as you don't have to account for IndexError allowing it to work for odd length as well.
If you want you can also add another parameter to let you specify the desired number of inner elements.
def func(lst, size = 2): # default of 2 it none specified
return [lst[i:i+size] for i in range(0, len(lst), size)]
There's a few hurdles in this problem. You want to iterate through the list without going past the end of the list and you need to deal with the case that list has an odd length. Here's one solution that works:
def foo(lst):
result = [[x,y] for [x,y] in zip(lst[0::2], lst[1::2])]
return result
In case this seems convoluted, let's break the code down.
Index slicing:
lst[0::2] iterates through lst by starting at the 0th element and proceeds in increments of 2. Similarly lst[1::2] iterates through starting at the 1st element (colloquially the second element) and continues in increments of 2.
Example:
>>> lst = (1,2,3,4,5,6,7)
>>> print(lst[0::2])
(1,3,5,7)
>>> print(lst[1::2])
(2,4,6)
zip: zip() takes two lists (or any iterable object for that matter) and returns a list containing tuples. Example:
>>> lst1 = (10,20,30, 40)
>>> lst2 = (15,25,35)
>>> prit(zip(lst1, lst2))
[(10,15), (20,25), (30,35)]
Notice that zip(lst1, lst2) has the nice property that if one of it's arguments is longer than the other, zip() stops zipping whenever the shortest iterable is out of items.
List comprehension: python allows iteration quite generally. Consider the statement:
>>> [[x,y] for [x,y] in zip(lst1,lst2)]
The interior bit "for [x,y] in zip(lst1,lst2)" says "iterate through all pairs of values in zip, and give their values to x and y". In the rest of the statement
"[[x,y] for [x,y] ...]", it says "for each set of values x and y takes on, make a list [x,y] to be stored in a larger list". Once this statement executes, you have a list of lists, where the interior lists are all possible pairs for zip(lst1,lst2)
Very Clear solution:
l = (1, 2, 3, 4, 5, 6)
l = iter(l)
w = []
for i in l:
sub = []
sub.append(i)
sub.append(next(l))
w.append(sub)
print w
>>> mylist = [ 1, 2, 3 , 5, 'd']
>>> print 'mylist[-1:2] = ',mylist[-1:2]
output is an empty list: []
I am testing around with lists and from what I have so far understood from tutorials made me think the output would be [d, 1, 2]
Can anyone please help me out why isn't that the output?
To understand why your code returns an empty list, it is important to note how accessing list elements in python works. Indexing with : on a python list (or slicing) works like this:
a_list[start:stop:step]
stop must always be greater than start for the list indexing/slicing to work.
When you access an element in the list by using a negative number as the index, say a_list[-1], python adds the length of the list to it and gives you what you want. For ex:
a_list = [1, 2, 3, 4, 5]
a_list[-1] == a[5 - 1]
So when you do mylist[-1:2] it actually means mylist[4:2]. This violates start < stop condition. And hence returns an empty list.
This is what you are looking for:
[mylist[-1]] + mylist[:2]
Slicing with negative numbers (see string slicing) does not make the list a ringbuffer. It's just another way of pointing to an element in the list. So, what you ask for is really: The list beginning at index 4 and ending at index 2. Doesn't make sense, hence you get nothing.
Slicing does not work that way in python. Instead, I suggest creating your own defined function, like below:
>>> def loop(list, start, end):
... array = []
... for k in range(start, end+1):
... array.append(list[k])
... return array
...
>>> x = [1, 2, 3, 4, 5]
>>> desired_output = [5, 1, 2, 3]
>>> loop(x, -1, 2)
[5, 1, 2, 3]
>>> loop(x, -1, 2) == desired_output
True
>>>
This loops over the list, and adds each value of the list to another array, which it then returns.
Or, you can add them together using the following code:
[x[-1]] + x[:2]