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I'm new to programming and am having some trouble with this exercise. The goal is to write a function that returns a list of matching items.
Items are defined by a tuple with a letter and a number and we consider item 1 to match item 2 if:
Both their letters are vowels (aeiou), or both are consonants
AND
The sum of their numbers is a multiple of 3
NOTE: The return list should not include duplicate matches --> (1,2) contains the same information as (2,1), the output list should only contain one of them.
Here's an example:
***input:*** [('a', 4), ('b', 5), ('c', 1), ('d', 3), ('e', 2), ('f',6)]
***output:*** [(0,4), (1,2), (3,5)]
Any help would be much appreciated!
from itertools import combinations
lst = [('a', 4), ('b', 5), ('c', 1), ('d', 3), ('e', 2), ('f',6)]
vowels = 'aeiou'
matched = [(i[0],j[0]) for (i,j) in combinations(enumerate(lst),2) if (i[1][0] in vowels) == (j[1][0] in vowels) and ((i[1][1] + j[1][1]) % 3 == 0)]
print(matched)
Sorry, I'm high enough rep to comment, but i'll edit / update once I can.
Im a little confused about the question, what is the purpose of the letters, should we be using their positon in the alphabet as their value? i.e a=0, b=1?
what are we comparing one tuple to?
Thanks
You can use itertools.combinations with enumerate to iterate all combinations and output indices. Combinations do not include permutations, so you will not see duplicates.
from itertools import combinations
lst = [('a', 4), ('b', 5), ('c', 1), ('d', 3), ('e', 2), ('f',6)]
def checker(lst):
vowels = set('aeiou')
for (idx_i, i), (idx_j, j) in combinations(enumerate(lst), 2):
if ((i[0] in vowels) == (j[0] in vowels)) and ((i[1] + j[1]) % 3 == 0):
yield idx_i, idx_j
res = list(checker(lst))
# [(0, 4), (1, 2), (3, 5)]
I have done something like this:
d = [('e', 0), ('f', 1), ('e', 0), ('f', 1)]
e = ['a']
d = [(n,j) for n,(i,j) in zip(e,d)]
d
[('a',0)]
I was just tryig to replace the equivalent tuple value with the array value, without changing the associated numbers. But the list only goes till the len of array e and not d. What I want to get as output is something like this:
d
[('a', 0), ('f', 1), ('e', 0), ('f', 1)]
Just add the unprocessed tail of d to the processed part:
[(n,j) for n,(i,j) in zip(e,d)] + d[len(e):]
#[('a', 0), ('f', 1), ('e', 0), ('f', 1)]
You can use itertools.zip_longest:
[(n or i, j) for n,(i,j) in itertools.zip_longest(e, d)]
Check the doc
If it's acceptable to mutate the original d list, I'd simply replace the first d tuples by iterating on e:
d = [('e', 0), ('f', 1), ('e', 0), ('f', 1)]
e = ['a']
for i, new_letter in enumerate(e):
d[i] = (new_letter, d[i][1])
print(d)
# [('a', 0), ('f', 1), ('e', 0), ('f', 1)]
Note that Python tuples are immutable. d[i][0] = new_letter would fail with the error:
TypeError: 'tuple' object does not support item assignment
The above code modifies the d list in place by replacing old tuples with new ones. It cannot modify the old tuples.
I think the problem is the zip function. The documentation says that (zip) "Returns an iterator of tuples, where the i-th tuple contains the i-th element from each of the argument sequences or iterables. The iterator stops when the shortest input iterable is exhausted."
[(n,j) for n,(i,j) in zip(e,d)] + d[len(e):] should do the trick
I having a list of tuple as describes below (This tuple is sorted in decreasing order of the second value):
from string import ascii_letters
myTup = zip (ascii_letters, range(10)[::-1])
threshold = 5.5
>>> myTup
[('a', 9), ('b', 8), ('c', 7), ('d', 6), ('e', 5), ('f', 4), ('g', 3), ('h', 2), \
('i', 1), ('j', 0)]
Given a threshold, what is the best possible way to discard all tuples having the second value less than this threshold.
I am having more than 5 million tuples and thus don't want to perform comparison tuple by tuple basis and consequently delete or add to another list of tuples.
Since the tuples are sorted, you can simply search for the first tuple with a value lower than the threshold, and then delete the remaining values using slice notation:
index = next(i for i, (t1, t2) in enumerate(myTup) if t2 < threshold)
del myTup[index:]
As Vaughn Cato points out, a binary search would speed things up even more. bisect.bisect would be useful, except that it won't work with your current data structure unless you create a separate key sequence, as documented here. But that violates your prohibition on creating new lists.
Still, you could use the source code as the basis for your own binary search. Or, you could change your data structure:
>>> myTup
[(0, 'a'), (1, 'b'), (2, 'c'), (3, 'd'), (4, 'e'), (5, 'f'),
(6, 'g'), (7, 'h'), (8, 'i'), (9, 'j')]
>>> index = bisect.bisect(myTup, (threshold, None))
>>> del myTup[:index]
>>> myTup
[(6, 'g'), (7, 'h'), (8, 'i'), (9, 'j')]
The disadvantage here is that the deletion may occur in linear time, since Python will have to shift the entire block of memory back... unless Python is smart about deleting slices that start from 0. (Anyone know?)
Finally, if you're really willing to change your data structure, you could do this:
[(-9, 'a'), (-8, 'b'), (-7, 'c'), (-6, 'd'), (-5, 'e'), (-4, 'f'),
(-3, 'g'), (-2, 'h'), (-1, 'i'), (0, 'j')]
>>> index = bisect.bisect(myTup, (-threshold, None))
>>> del myTup[index:]
>>> myTup
[(-9, 'a'), (-8, 'b'), (-7, 'c'), (-6, 'd')]
(Note that Python 3 will complain about the None comparison, so you could use something like (-threshold, chr(0)) instead.)
My suspicion is that the linear time search I suggested at the beginning is acceptable in most circumstances.
Here's an exotic approach that wraps the list in a list-like object before performing bisect.
import bisect
def revkey(items):
class Items:
def __getitem__(self, index):
assert 0 <= index < _len
return items[_max-index][1]
def __len__(self):
return _len
def bisect(self, value):
return _len - bisect.bisect_left(self, value)
_len = len(items)
_max = _len-1
return Items()
tuples = [('a', 9), ('b', 8), ('c', 7), ('d', 6), ('e', 5), ('f', 4), ('g', 3), ('h', 2), ('i', 1), ('j', 0)]
for x in range(-2, 12):
assert len(tuples) == 10
t = tuples[:]
stop = revkey(t).bisect(x)
del t[stop:]
assert t == [item for item in tuples if item[1] >= x]
Maybe a bit faster code than of #Curious:
newTup=[]
for tup in myTup:
if tup[1]>threshold:
newTup.append(tup)
else:
break
Because the tuples are ordered, you do not have to go through all of them.
Another possibility would also be, to use bisection, and find the index i of last element, which is above threshold. Then you would do:
newTup=myTup[:i]
I think the last method would be the fastest.
Given the number of tuples you're dealing with, you may want to consider using NumPy.
Define a structured array like
my_array= np.array(myTup, dtype=[('f0',"|S10"), ('f1',float)])
You can access the second elements of your tuples with myarray['f1'] which gives you a float array. Youcan know use fancy indexing techniques to filter the elements you want, like
my_array[myarray['f1'] < threshold]
keeping only the entries where your f1 is less than your threshold..
You can also use itertools e.g.
from itertools import ifilter
iterable_filtered = ifilter(lambda x : x[1] > threshold, myTup)
If you wanted an iterable filtered list or just:
filtered = filter(lambda x: x[1] > threshold, myTup)
to go straight to a list.
I'm not too familiar with the relative performance of these methods and would have to test them (e.g. in IPython using %timeit).
Am looking for a clean pythonic way of doing the following
I have a list of tuples say :
[(1,'a'), (1,'b'), (1,'c'), (2, 'd'), (5, 'e'), (5, 'f')]
I want to make a new list which discards tuples whose first key has been seen before. So the o/p for the above would be:
[(1,'c'), (2,'d'), (5, 'f')]
Thanks!
A simple way would be creating a dictionary, since it will only keep the last element with the same key:
In [1]: l = [(1,'a'), (1,'b'), (1,'c'), (2, 'd'), (5, 'e'), (5, 'f')]
In [2]: dict(l).items()
Out[2]: [(1, 'c'), (2, 'd'), (5, 'f')]
Update: As #Tadeck mentions in his comment, since the order of dictionary items is not guaranteed, you probably want to use an ordered dictionary:
from collections import OrderedDict
newl = OrderedDict(l).items()
If you actually want to keep the first tuple with the same key (and not the last, your question is ambiguous), then you could reverse the list first, add it do the dictionary and reverse the output of .items() again.
Though in that case there are probably better ways to accomplish this.
Using unique_everseen from itertools docs
from itertools import ifilterfalse
def unique_everseen(iterable, key=None):
"List unique elements, preserving order. Remember all elements ever seen."
# unique_everseen('AAAABBBCCDAABBB') --> A B C D
# unique_everseen('ABBCcAD', str.lower) --> A B C D
seen = set()
seen_add = seen.add
if key is None:
for element in ifilterfalse(seen.__contains__, iterable):
seen_add(element)
yield element
else:
for element in iterable:
k = key(element)
if k not in seen:
seen_add(k)
yield element
a = [(1,'a'), (1,'b'), (1,'c'), (2, 'd'), (5, 'e'), (5, 'f')]
print list(unique_everseen(a,key=lambda x: x[0]))
Yielding
[(1, 'a'), (2, 'd'), (5, 'e')]
a nifty trick for one liner fetishists that keeps the order in place (I admit it's not very readable but you know...)
>>> s = [(1,'a'), (1,'b'), (1,'c'), (2, 'd'), (5, 'e'), (5, 'f')]
>>> seen = set()
>>> [seen.add(x[0]) or x for x in s if x[0] not in seen]
[(1, 'a'), (2, 'd'), (5, 'e')]
My program is not doing a great job. In a loop, data from each processor (list of tuple) are gathered into the master processor that needs to clean it by removing similar element.
I found a lot of interesting clue on internet and especially in this site about union of list. However, i have not managed to apply it to my problem.
My aim is to get rid of tuple whose its two last element are similar to another tuple in the list . for example:
list1=[[a,b,c],[d,e,f],[g,h,i]]
list2=[[b,b,c],[d,e,a],[k,h,i]]
the result should be:
final=[[a,b,c],[d,e,f],[g,h,i],[d,e,a]]
Right now I'm using loops and break but I'm hoping to make this process faster.
here is what my code looks like (result and temp are the lists I want to get union from)
on python2.6.
for k in xrange(len(temp)):
u=0
#index= next((j for j in xrange(lenres) if temp[k][1:3] == result[j][1:3]),None)
for j in xrange(len(result)):
if temp[k][1:3] == result[j][1:3]:
u=1
break
if u==0:
#if index is None:
result.append([temp[k][0],temp[k][1],temp[k][2]])
Thanks for your help
Herve
Below is our uniques function. It takes arguments l (list) and f (function), returns list with duplicates removed (in the same order). Duplicates are defined by: b is duplicate of a iff f(b) == f(a).
def uniques(l, f = lambda x: x):
return [x for i, x in enumerate(l) if f(x) not in [f(y) for y in l[:i]]]
We define lastTwo as follows:
lastTwo = lambda x: x[-2:]
For your problem we use it as follows:
>>> list1
[('a', 'b', 'c'), ('d', 'e', 'f'), ('g', 'h', 'i')]
>>> list2
[('b', 'b', 'c'), ('d', 'e', 'a'), ('k', 'h', 'i')]
>>> uniques(list1+list2, lastTwo)
[('a', 'b', 'c'), ('d', 'e', 'f'), ('g', 'h', 'i'), ('d', 'e', 'a')]
If the usecase you describe comes up a lot you may want to define
def hervesMerge(l1, l2):
return uniques(l1+l2, lambda x: x[-2:])
Identity is our default f but it can be anything (so long as it is defined for all elements of the list, since they can be of any type).
f can be sum of a list, odd elements of a list, prime factors of an integer, anything. (Just remember that if its injective theres no point! Add by constant, linear functions, etc will work no differently than identity bc its f(x) == f(y) w/ x != y that makes the difference)
>>> list1
[(1, 2, 3, 4), (2, 5), (6, 2, 2), (3, 4), (8, 3), (1, 1, 1, 1, 1, 1, 1, 1, 1, 1)]
>>> uniques(list1, sum)
[(1, 2, 3, 4), (2, 5), (8, 3)]
>>> uniques(list1, lambda x: reduce(operator.mul, x)) #product
[(1, 2, 3, 4), (2, 5), (3, 4), (1, 1, 1, 1, 1, 1, 1, 1, 1, 1)]
>>> uniques([1,2,3,4,1,2]) #defaults to identity
[1, 2, 3, 4]
You seemed concerned about speed, but my answer really focused on shortness/flexibility without significant (or any?) speed improvment. For bigger lists where speed is z concern, you want to take advantage of hashable checks and the fact that list1 and list2 are known to have no duplicates
>>> s = frozenset(i[-2:] for i in list1)
>>> ans = list(list1) #copy list1
>>> for i in list2:
if i[-2:] not in s: ans.append(i)
>>> ans
[('a', 'b', 'c'), ('d', 'e', 'f'), ('g', 'h', 'i'), ('d', 'e', 'a')]
OR allowing disordering
>>> d = dict()
>>> for i in list2 + list1:
d[i[-2:]] = i
>>> d.values()
[('d', 'e', 'f'), ('a', 'b', 'c'), ('g', 'h', 'i'), ('d', 'e', 'a')]
--Edit--
You should always be able to avoid un-pythonic looping like you post in your question. Here is your exact code with the loops changed:
for k in temp:
u=0
for j in result:
if k[1:3] == j[1:3]:
u=1
break
if u==0:
#if index is None:
result.append([k[0],k[1],k[2]]) // k
result and temp are iterable, and for anything iterable you can put it directly in the for loop without eanges. If for some reason you explicitly need the index (this is not such a case, but I have one above) you can use enumerate.
Here's a simple solution using a set:
list1=[('a','b','c'),('d','e','f'),('g','h','i')]
list2=[('b','b','c'),('d','e','a'),('k','h','i')]
set1 = set([A[1:3] for A in list1])
final = list1 + [A for A in list2 if A[1:3] not in set1]
However, if your list1 and list2 aren't actually made of tuples, then you will have to put tuple() around A[1:3].