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How to convert string to list inside of list?

I don't know how to convert string to list inside of list. Split method seems the most popular, but it doesn't work.
I have a list:
corpus = ['[(0, 1), (1, 2), (2, 1)]']
I want to convert string to list like that:
corpus = [[(0, 1), (1, 2), (2, 1)]]
How I can do it?

Use ast.literal_eval, a safer alternative to eval:
import ast
corpus = ['[(0, 1), (1, 2), (2, 1)]']
corpus = [ast.literal_eval(item) for item in corpus]
print(corpus)
# [[(0, 1), (1, 2), (2, 1)]]

Hi I think this simple code will help.
import re
corpus = ['[(0, 1), (1, 2), (2, 1)]']
corpus[0] = re.findall(r'\(\d, \d\)', corpus[0])
temp_lst1 = []
for i in corpus[0]:
temp_lst2 = []
for j in i:
try:
if type(int(j)) == int:
temp_lst2.append(int(j))
except ValueError:
pass
temp_lst1.append(tuple(temp_lst2))
corpus[0] = temp_lst1
print(corpus)

How to sort a list of tuples which contain two integer element firstly by its second, then by its first element descending in Python?

I've been using loops in order to solve my problem but my brain has stopped. I have a input as a list which contains couple of tuples like this:
input_list = [(1, 2), (2, 2), (3, 2), (5, 3), (7, 3), (4, 4)]
I would like to mix this tuples. Here are the steps:
Tuples have to be sorted in ascending by its second element for every time. ((X, 2), (X, 3), (X, 4), (X, 2), (X, 3), (X, 2))
First element of the tuple must be the smallest value. ((1, 2), (5, 3), (4, 4), (2, 2), (7, 3), (3, 2))
Here is the output I would like to achieve:
output_list = [(1, 2), (5, 3), (4, 4), (2, 2), (7, 3), (3, 2)]
Can someone help me to find the algorithm to mix and convert to input list to the output list?

Sort the input by the second, then the first element. For each tuple in the sorted list, iterate a list of "slots" (lists) and add the tuple to the first slot whose last item is less than the tuple. If there's no such slot, create a new slot. Finally, join the slots into one list:
slots = []
for p in sorted(INPUT, key=lambda p: p[::-1]):
for s in slots:
if s[-1][1] < p[1]:
s.append(p)
break
else:
slots.append([p])
result = sum(slots, [])

You could use roundrobin from the itertools recipes. The function group_values sorts the data so that it can be used as an input for roundrobin.
from collections import defaultdict
from itertools import cycle, islice
def roundrobin(*iterables):
"roundrobin('ABC', 'D', 'EF') --> A D E B F C"
# Recipe credited to George Sakkis
num_active = len(iterables)
nexts = cycle(iter(it).__next__ for it in iterables)
while num_active:
try:
for next_ in nexts:
yield next_()
except StopIteration:
# Remove the iterator we just exhausted from the cycle.
num_active -= 1
nexts = cycle(islice(nexts, num_active))
def group_values(data):
result = defaultdict(list)
# sort by the second element first
for entry in sorted(data, key=lambda x: (x[1], x[0])):
result[entry[1]].append(entry)
return result
def main():
input_list = [(1, 2), (2, 2), (3, 2), (5, 3), (7, 3), (4, 4)]
grouped_lists = group_values(input_list)
result = list(roundrobin(*grouped_lists.values()))
print(result)
if __name__ == '__main__':
main()
group_values creates a dictionary that looks like this: {2: [(1, 2), (2, 2), (3, 2)], 3: [(5, 3), (7, 3)], 4: [(4, 4)]}. The function takes advantage of the fact that in modern versions of Python, entries in the dictionaries retain the order in which they were added.
The result is [(1, 2), (5, 3), (4, 4), (2, 2), (7, 3), (3, 2)], even if you shuffle the input list.
You might want to use a debugger or add some print functions in the code if you want to see what's happening.

Problem with using sets and issubset in Python

I'm not clear about how my sets needs to be written. The answer I'm getting is False, which means I'm using the issubset type wrong or my sets are not written correctly.
some_cords = {(1, 2), (2, 2), (3, 2)}
line_cords = {((1, 2), (2, 2), (3, 2)), ((1, 3), (2, 3), (3, 3)), ((2, 1), (2, 2), (2, 3)),
((2, 4), (2, 3), (2, 2))}
print(some_cords.issubset(line_cords))
print(any(some_cords in k for k in line_cords))
>>False
>>False
I've tried these two methods but why they're False, I'm not sure. I've also tried writing some_cords and line_cords as tuples rather than sets, i.e.
some_cords = ((1, 2), (2, 2), (3, 2)) but I'm getting the same result. I'm running Python 3.10 in IntelliJ IDE. Thanks for any help.

As written, some_cords is a set with 3 elements, each one is a tuple: (1, 2), (2, 2) and (3, 2). On the other hand, line_cords is a set of 4 tuples, each tuple being a tuple of length 3.
Adding parenthesis around the tuples of some_cords will turn it into a set with one element similar to the ones of line_cords:
some_cords = {((1, 2), (2, 2), (3, 2))}
and the first test will return True
As for the second test (using in), some_cords is a set, and each element of line_cords is a tuple, so it will always return False. This test would be OK if some_cords was a tuple: you can achieve this removing the braces, but then you can not use issubset.

How to remove duplicate from list of tuple when order is important

I have seen some similar answers, but I can't find something specific for this case.
I have a list of tuples:
[(5, 0), (3, 1), (3, 2), (5, 3), (6, 4)]
What I want is to remove tuples from this list only when first element of tuple has occurred previously in the list and the tuple which remains should have the smallest second element.
So the output should look like this:
[(5, 0), (3, 1), (6, 4)]

Here's a linear time approach that requires two iterations over your original list.
t = [(5, 0), (3, 1), (3, 2), (5, 3), (6, 4)] # test case 1
#t = [(5, 3), (3, 1), (3, 2), (5, 0), (6, 4)] # test case 2
smallest = {}
inf = float('inf')
for first, second in t:
if smallest.get(first, inf) > second:
smallest[first] = second
result = []
seen = set()
for first, second in t:
if first not in seen and second == smallest[first]:
seen.add(first)
result.append((first, second))
print(result) # [(5, 0), (3, 1), (6, 4)] for test case 1
# [(3, 1), (5, 0), (6, 4)] for test case 2

Here is a compact version I came up with using OrderedDict and skipping replacement if new value is larger than old.
from collections import OrderedDict
a = [(5, 3), (3, 1), (3, 2), (5, 0), (6, 4)]
d = OrderedDict()
for item in a:
# Get old value in dictionary if exist
old = d.get(item[0])
# Skip if new item is larger than old
if old:
if item[1] > old[1]:
continue
#else:
# del d[item[0]]
# Assign
d[item[0]] = item
list(d.values())
Returns:
[(5, 0), (3, 1), (6, 4)]
Or if you use the else-statement (commented out):
[(3, 1), (5, 0), (6, 4)]

Seems to me that you need to know two things:
The tuple that has the smallest second element for each first element.
The order to index each first element in the new list
We can get #1 by using itertools.groupby and a min function.
import itertools
import operator
lst = [(3, 1), (5, 3), (5, 0), (3, 2), (6, 4)]
# I changed this slightly to make it harder to accidentally succeed.
# correct final order should be [(3, 1), (5, 0), (6, 4)]
tmplst = sorted(lst, key=operator.itemgetter(0))
groups = itertools.groupby(tmplst, operator.itemgetter(0))
# group by first element, in this case this looks like:
# [(3, [(3, 1), (3, 2)]), (5, [(5, 3), (5, 0)]), (6, [(6, 4)])]
# note that groupby only works on sorted lists, so we need to sort this first
min_tuples = {min(v, key=operator.itemgetter(1)) for _, v in groups}
# give the best possible result for each first tuple. In this case:
# {(3, 1), (5, 0), (6, 4)}
# (note that this is a set comprehension for faster lookups later.
Now that we know what our result set looks like, we can re-tackle lst to get them in the right order.
seen = set()
result = []
for el in lst:
if el not in min_tuples: # don't add to result
continue
elif el not in seen: # add to result and mark as seen
result.append(el)
seen.add(el)

This will do what you need:
# I switched (5, 3) and (5, 0) to demonstrate sorting capabilities.
list_a = [(5, 3), (3, 1), (3, 2), (5, 0), (6, 4)]
# Create a list to contain the results
list_b = []
# Create a list to check for duplicates
l = []
# Sort list_a by the second element of each tuple to ensure the smallest numbers
list_a.sort(key=lambda i: i[1])
# Iterate through every tuple in list_a
for i in list_a:
# Check if the 0th element of the tuple is in the duplicates list; if not:
if i[0] not in l:
# Add the tuple the loop is currently on to the results; and
list_b.append(i)
# Add the 0th element of the tuple to the duplicates list
l.append(i[0])
>>> print(list_b)
[(5, 0), (3, 1), (6, 4)]
Hope this helped!

Using enumerate() and list comprehension:
def remove_if_first_index(l):
return [item for index, item in enumerate(l) if item[0] not in [value[0] for value in l[0:index]]]
Using enumerate() and a for loop:
def remove_if_first_index(l):
# The list to store the return value
ret = []
# Get the each index and item from the list passed
for index, item in enumerate(l):
# Get the first number in each tuple up to the index we're currently at
previous_values = [value[0] for value in l[0:index]]
# If the item's first number is not in the list of previously encountered first numbers
if item[0] not in previous_values:
# Append it to the return list
ret.append(item)
return ret
Testing
some_list = [(5, 0), (3, 1), (3, 2), (5, 3), (6, 4)]
print(remove_if_first_index(some_list))
# [(5, 0), (3, 1), (6, 4)]

I had this idea without seeing the #Anton vBR's answer.
import collections
inp = [(5, 0), (3, 1), (3, 2), (5, 3), (6, 4)]
od = collections.OrderedDict()
for i1, i2 in inp:
if i2 <= od.get(i1, i2):
od.pop(i1, None)
od[i1] = i2
outp = list(od.items())
print(outp)

How to generate list of tuples relating records

I need to generate a list from the list of tuples:
a = [(1,2), (1,3), (2,3), (2,5), (2,6), (3,4), (3,6), (4,7), (5 6), (5,9), (5,10), (6,7)
(6.10) (6.11) (7.8) (7.12) (8.12) (9.10) (10.11)]
The rule is:
- I have a record from any (begin = random.choice (a))
- Items from the new list must have the following relationship:
the last item of each tuple in the list must be equal to the first item of the next tuple to be inserted.
Example of a valid output (starting by the tuple (3.1)):
[(3, 1), (1, 2), (2, 3), (3, 4), (4, 7), (7, 8), (8, 12), (12, 7), (7, 6), (6, 2), (2, 5), (5, 6), (6, 10), (10, 5) (5, 9), (9, 10), (10, 11), (11, 6), (6, 3)]
How can I do this? Its make using list comprehensions?
Thanks!

Here, lisb will be populated with tuples in the order that you seek. This is, of course, if lisa provides appropriate tuples (ie, each tuple has a 1th value matching another tuple's 0th value). Your sample list will not work, regardless of the implementation, because all the values don't match up (for example, there is no 0th element with 12, so that tuple can't be connected forward to any other tuple)...so you should come up with a better sample list.
Tested, working.
import random
lisa = [(1, 2), (3, 4), (2, 3), (4, 0), (0, 9), (9, 1)]
lisb = []
current = random.choice(lisa)
while True:
lisa.remove(current)
lisb.append(current)
current = next((y for y in lisa if y[0] == current[1]), None)
if current == None:
break
print lisb
If you don't want to delete items from lisa, just slice a new list.

As a generator function:
def chained_tuples(x):
oldlist = x[::]
item = random.choice(oldlist)
oldlist.remove(item)
yield item
while oldlist:
item = next(next_item for next_item in oldlist if next_item[0] == item[1])
oldlist.remove(item)
yield item
As noted, you'll get an incomplete response if your list isn't actually chainable all the way through, like your example list.

Just to add another way of solving this problem:
import random
from collections import defaultdict
lisa = [(1, 2), (3, 4), (2, 3), (4, 0), (0, 9), (9, 1)]
current_start, current_end = lisa[random.randint(0, len(lisa) - 1)]
starts = defaultdict(list)
lisb = [(current_start, current_end)]
for start, end in lisa:
starts[start].append(end)
while True:
if not starts[current_end]:
break
current_start, current_end = current_end, starts[current_end].pop()
lisb.append((current_start, current_end))
Note: You have to make sure lisa is not empty.

I think all of the answers so far are missing the requirement (at least based on your example output) that the longest chain be found.
My suggested solution is to recursively parse all possible chains that can be constructed, and return the longest result. The function looks like this:
def generateTuples(list, offset, value = None):
if value == None: value = list[offset]
list = list[:offset]+list[offset+1:]
res = []
for i,(a,b) in enumerate(list):
if value[1] in (a,b):
if value[1] == a:
subres = generateTuples(list, i, (a,b))
else:
subres = generateTuples(list, i, (b,a))
if len(subres) > len(res):
res = subres
return [value] + res
And you would call it like this:
results = generateTuples(a, 1, (3,1))
Producing the list:
[(3, 1), (1, 2), (2, 3), (3, 4), (4, 7), (7, 8), (8, 12), (12, 7), (7, 6),
(6, 2), (2, 5), (5, 6), (6, 10), (10, 5), (5, 9), (9, 10), (10, 11),
(11, 6), (6, 3)]
The first parameter of the function is the source list of tuples, the second parameter is the offset of the first element to use, the third parameter is optional, but allows you to override the value of the first element. The latter is useful when you want to start with a tuple in its reversed order as you have done in your example.