慕标琳琳
由于您是从上到下，然后从左到右进行视觉搜索，因此此代码要简单得多并提供正确的结果。它基本上相当于视觉扫描，通过检查每个“y=n”位置的所有元组，然后根据第二个数字（从左到右）对任何“y=n”元组进行排序。为了与笛卡尔数系统更加一致，我已将图表上的点转换为 (x,y) 坐标，其中 X 为正（向右增加）和 y 为负（随着它们下降而减少）。d = {(2,-4):1, (5,-3):2, (4,-1):3, (1,-1):4, (2,-2):5, (3,-1):6, (1,-5):7}l = [(2,-4), (5,-3), (4,-1), (1,-1), (2,-2), (3,-1), (1,-5)]results = []# Use the length of the list. Its more than needed, but guarantees enough loopsfor y in range(0, -len(l), -1): # For ONLY the items found at the specified y coordinate temp_list = [] for i in l: # Loop through ALL the items in the list if i[1] == y: # If tuple is at this "y" coordinate then... temp_list.append(i) # ... append it to the temp list # Now sort the list based on the "x" position of the coordinate temp_list = sorted(temp_list, key=lambda x: x[0]) results += temp_list # And just append it to the final result list# Final TUPLES in orderprint(results)# If you need them correlated to their original numbersby_designator_num = []for i in results: # The the first tupele value by_designator_num.append(d[i]) # Use the tuple value as the key, to get the original designator number from the original "d" dictionaryprint(by_designator_num)或者如果你想要它更快更紧凑d = {(2,-4):1, (5,-3):2, (4,-1):3, (1,-1):4, (2,-2):5, (3,-1):6, (1,-5):7}l = [(2,-4), (5,-3), (4,-1), (1,-1), (2,-2), (3,-1), (1,-5)]results = []for y in range(0, -len(l), -1): results += sorted([i for i in l if i[1] == y ], key=lambda x: x[0])print(results)by_designator_num = [d[i] for i in results]print(by_designator_num)输出：[(1, -1), (3, -1), (4, -1), (2, -2), (5, -3), (2, -4), (1, -5)][4, 6, 3, 5, 2, 1, 7]
慕雪6442864
这基于对元组的第一个坐标进行排序，然后按元组的第二个坐标对代码进行排序。即按字母顺序，“Aa”，然后是“Ab”，然后是“Ba”，然后是“Bb”。更字面意思是 (1,1)、(1,2)、(2,1)、(2,2) 等。如果（且仅当）与 #7 关联的元组值对在您的问题中实际上是乱序的（并且实际上应该在 #3 和 #5 之间），这将起作用。如果不是这种情况，请参阅我的其他答案。# Make it a dictionary, with the VALUETUPLES as the KEYS, and the designator as the valued = {(1,1):4, (1,3):6, (1,4):3, (2,2):5, (3,5):2, (4,2):1,(1,5):7}# ALSO make a list of just the value tuplesl = [ (1,1), (1,3), (1,4), (2,2), (3,5), (4,2), (1,5)]# Sort the list by the first element in each tuple. ignoring the secondnew = sorted(l, key=lambda x: x[0])# Create a new dictionary, basically for temp sortingnew_d = {}# This iterates through the first sorted list "new"# and creates a dictionary where the key is the first number of value tuplescount = 0# The extended range is because we don't know if any of the Tuple Values share any same numbersfor r in range(0, len(new)+1,1): count += 1 new_d[r] = [] for item in new: if item[0] == r: new_d[r].append(item)print(new_d) # So it makes sense# Make a final list to capture the rdered TUPLES VALUESfinal_list = []# Go through the same rage as abovefor r in range(0, len(new)+1,1): _list = new_d[r] # Grab the first list item from the dic. Order does not matter here if len(_list) > 0: # If the list has any values... # Sort that list now by the SECOND tuple value _list = sorted(_list, key=lambda x: x[1]) # Lists are ordered. So we can now just tack that ordered list onto the final list. # The order remains for item in _list: final_list.append(item)# This is all the tuple values in orderprint(final_list)# If you need them correlated to their original numbersby_designator_num = []for i in final_list: # The the first tupele value by_designator_num.append(d[i]) # Use the tuple value as the key, to get the original designator number from the original "d" dictionaryprint(by_designator_num)输出：[(1, 1), (1, 3), (1, 4), (1, 5), (2, 2), (3, 5), (4, 2)][4, 6, 3, 7, 5, 2, 1]