Oscar Benjamin wrote: > On 1 August 2018 at 21:38, Roger Lea Scherer <rls...@gmail.com> wrote: >> >> I'm trying to get a list of tuples to be a float, a numerator, and a >> denominator for all the fractions: halves, thirds, fourths etc up to >> ninths. 1/2 returns the same float as 2/4, 3/6, 4/8. I would like to keep >> only the 1/2. When I try (line 18) to "pop" from the list I get a >> "TypeError: >> integer argument expected, got float". When I try (line 18) to "remove" >> from the list, nothing happens: nothing is removed and I do not receive >> an error message. >> >> What do you think is a good way to solve this? >> >> Thank you as always. >> >> import math >> >> fractions = [(0, 0, 0)] >> >> for i in range(1, 10): >> for j in range(1, 10): >> if i < j: >> x = i/j >> if x not in fractions: >> fractions.append((x, i, j)) >> sortedFrac = sorted(fractions) >> >> print(sortedFrac) >> >> for i in range(len(sortedFrac)): >> try: >> if sortedFrac[i][0] == sortedFrac[i-1][0]: # so if the float >> equals >> the previous float >> sortedFrac.pop(sortedFrac[i][0]) # remove the >> second >> float >> else: >> sortedFrac.append(sortedFrac[i][0]) >> except ValueError: >> continue > > Comparing floats for equality can be flakey. Sometimes two floats that > should be equal will not compare equal e.g.: > >>>> 0.01 + 0.1 - 0.1 == 0.01 > False
Do you know if there's a way to construct an example where i/k != (n*i)/(n*k) with preferrably small integers i, k, and n? Python's integer division algorithm defeats my naive attempts ;) I had to resort to i/k == i/(k + 1) >>> k = 10**18 >>> 1/k == 1/(k+1) True > This happens in this case because of intermediate rounding errors. I > don't think that should affect you since you are doing precisely one > floating point operation i/j to calculate each of your floats but > unless you have a very solid understanding of binary floating point I > would recommend that you shouldn't compare floating point values as in > a==b. > > For this particular problem I would use integer arithmetic or I would > use the fractions module. Doing this with integers you should > normalise the numerator and denominator by dividing out their GCD. The > fractions module takes care of this for you internally. > https://docs.python.org/3.7/library/fractions.html > > Otherwise in general to remove duplicates you would be better of with > a set rather than a list. If you only put x in the set and not i and j > then the set will automatically take care of duplicates: > https://docs.python.org/3.7/tutorial/datastructures.html#sets And if you want to keep i and j you can use a dict: fractions = {} for i in range(1, 10): for j in range(1, 10): if i < j: x = i/j if x not in fractions: fractions[x] = x, i, j sorted_unique_fractions = sorted(fractions.values()) _______________________________________________ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
