How Do I Extract Part Of A Tuple That's Duplicate As Key To A Dictionary, And Have The Second Part Of The Tuple As Value?

I'm pretty new to Python and Qgis, right now I'm just running scripts but I my end-goal is to create a plugin. Here's the part of the code I'm having problems with: import math la

Solution 1:

Maybe this is what you need?

dictionary = {}
for i indict:
    for j indict:
        c = set(dict[i]).intersection(set(dict[j]))
        iflen(c) == 1:
            # ok, so now we know, that exactly one tuple exists in both# sets at the same time, but this one will be the key to new dictionary# we need the second tuple from the set to become value for this new key# so we can subtract the key-tuple from set to get the other tuple
            d = set(dict[i]).difference(c)
            # Now we need to get tuple back from the set# by doing list(c) we get list# and our tuple is the first element in the list, thus list(c)[0]
            c = list(c)[0]
            dictionary[c] = list(d)[0]
        else: pass

This code attaches only one tuple to the key in dictionary. If you want multiple values for each key, you can modify it so that each key would have a list of values, this can be done by simply modifying:

# some_value cannot be a set, it can be obtained with c = list(c)[0]
key = some_value
dictionary.setdefault(key, [])
dictionary[key].append(value)

So, the correct answer would be:

dictionary ={}for i in a:for j in a:c= set(a[i]).intersection(set(a[j]))if len(c)==1:
                d = set(a[i]).difference(c)c=list(c)[0]
                value =list(d)[0]ifcin dictionary and value not in dictionary[c]:
                    dictionary[c].append(value)
                elif c not in dictionary:
                    dictionary.setdefault(c,[])
                    dictionary[c].append(value)else: pass

Solution 2:

See this code :

dict={0L: [(355277,6.68901e+06), (355385,6.68906e+06)],1L: [(355238,6.68909e+06), (355340,6.68915e+06)],2L: [(355340,6.68915e+06), (355452,6.68921e+06)],3L: [(355340,6.68915e+06), (355364,6.6891e+06)],4L: [(355364,6.6891e+06), (355385,6.68906e+06)],5L: [(355261,6.68905e+06), (355364,6.6891e+06)],6L: [(355364,6.6891e+06), (355481,6.68916e+06)],7L: [(355385,6.68906e+06), (355501,6.68912e+06)]}dictionary= {}
list=[]for item in dict :list.append(dict[0])list.append(dict[1])b= []

[b.append(x)forcinlistforxincifxnotinb]
printb# or set(b)res={}for elm in b :lst=[]for item in dict :ifdict[item][0]==elm :lst.append(dict[item][1])elifdict[item][1]==elm :lst.append(dict[item][0])res[elm]=lstprintres