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Get route from unordered set of tuples


Get route from unordered set of tuples

By : user2185802
Date : November 22 2020, 04:01 AM
I wish this help you I guess this is just an example to learn how to do certain things with Python? Have a look at the code below for a very basic approach how you would be able to do it. It's not optimized code, though.
code :
# Define a route by the steps you have to take from one origin to the
# new destination. These steps are provided as a list of tuples where
# the first element is the origin and the second element is the destination.
steps = [(0,10), (11,0), (10,11)]

# Start with origin 0
route = [0]

# Repeat until no steps left to take
while len(steps) > 0:
    # The former destination is the last element of the current route
    former_destination = route[-1]
    # Browse through all remaining steps
    for i, (origin, destination) in enumerate(steps):
        # Skip this step if its origin and the former destination don't match
        if origin != former_destination:
            continue
        # Remove the current step
        steps.pop(i)
        # Add new destination
        route.append(destination)
        break

print(route)
[0, 10, 11, 0]


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Building an unordered map with tuples as keys

Building an unordered map with tuples as keys


By : Piyush Jain
Date : March 29 2020, 07:55 AM
this one helps. You need a bit of front matter. Because of the underlying implementation of boost::tuples::tuple, make Edge a structure to allow the overloads to resolve correctly. Otherwise, you'll get no matches for
boost::hash_value(const Edge &) operator==(const Edge &, const Edge &)
code :
struct Edge {
  Edge(double x1, double x2, double x3, double x4)
    : tuple(x1,x2,x3,x4) {}
  boost::tuples::tuple<double, double, double, double> tuple;
};

// XXX: less than ideal implementation!
bool operator==(const Edge &a, const Edge &b)
{
  return a.tuple.get<0>() == b.tuple.get<0>() &&
         a.tuple.get<1>() == b.tuple.get<1>() &&
         a.tuple.get<2>() == b.tuple.get<2>() &&
         a.tuple.get<3>() == b.tuple.get<3>();
}

// XXX: me too!
std::size_t hash_value(const Edge &e)
{
  std::size_t seed = 0;
  boost::hash_combine(seed, e.tuple.get<0>());
  boost::hash_combine(seed, e.tuple.get<1>());
  boost::hash_combine(seed, e.tuple.get<2>());
  boost::hash_combine(seed, e.tuple.get<3>());
  return seed;
}

typedef boost::unordered_map< Edge, int > EdgeMap;
MySQL remove distinct unordered tuples

MySQL remove distinct unordered tuples


By : Marcelo Augusto Cost
Date : March 29 2020, 07:55 AM
hop of those help? In my result set, I have records such as (a,b), (b,a) I only need to show distinct pair, not taking into consideration the order. , Change AND A.model != B.model to AND A.model < B.model.
Converting unordered list of tuples to pandas DataFrame

Converting unordered list of tuples to pandas DataFrame


By : Bert
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further Thank you for your responses! I ended up doing a completely different workaround as follows:
I checked the documentation to see all possible parse_tags from usaddress, created a DataFrame with all possible tags as columns, and one other column with the extracted addresses. Then I proceeded to parse and extract information from the columns using regex. Code below!
code :
parse_tags = ['Recipient','AddressNumber','AddressNumberPrefix','AddressNumberSuffix',
'StreetName','StreetNamePreDirectional','StreetNamePreModifier','StreetNamePreType',
'StreetNamePostDirectional','StreetNamePostModifier','StreetNamePostType','CornerOf',
'IntersectionSeparator','LandmarkName','USPSBoxGroupID','USPSBoxGroupType','USPSBoxID',
'USPSBoxType','BuildingName','OccupancyType','OccupancyIdentifier','SubaddressIdentifier',
'SubaddressType','PlaceName','StateName','ZipCode']

addr = ['123 Pennsylvania Ave NW Washington DC 20008', 
        '652 Polk St San Francisco, CA 94102', 
        '3711 Travis St #800 Houston, TX 77002']

df = pd.DataFrame({'Addresses': addr})
pd.concat([df, pd.DataFrame(columns = parse_tags)])
df['Info'] = df['Addresses'].apply(lambda x: str(usaddress.parse(x)))
for colname in parse_tags:
    df[colname] = df['Info'].apply(lambda x: re.findall("\('(\S+)', '{}'\)".format(colname), x)[0] if re.search(
    colname, x) else "")
Remove duplicate unordered tuples from list

Remove duplicate unordered tuples from list


By : Erik Fjelldal
Date : March 29 2020, 07:55 AM
I wish this helpful for you You could use combinations_with_replacement
code :
import itertools
pairs = list(itertools.combinations_with_replacement(range(3), 2))
print(pairs)

>>> [(0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2)]
Difference between two list of unordered tuples

Difference between two list of unordered tuples


By : Sheo Shankar
Date : March 29 2020, 07:55 AM
it helps some times I have two lists with tuples as shown below , You can use sets and come back to list:
code :
>>> list(set([tuple(sorted(x)) for x in a]) - set([tuple(sorted(x)) for x in b]))
[('d', 'e', 'h')]
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