我有這個問題。我需要訂購這點 1-7 1(4,2), 2(3, 5), 3(1,4), 4(1,1), 5(2,2), 6(1,3), 7(1,5)并得到這個結果 4 , 6 , 3 , 5 , 2 , 1 , 7。我正在使用 python 腳本進行帶有 x 引用的排序并且沒問題，但是 y 中的排序是錯誤的。我試過 sorted(dicts,key=itemgetter(1,2))有人可以幫助我嗎？
3 回答
躍然一笑
TA貢獻1826條經驗 獲得超6個贊
嘗試這個：
sorted(dicts,key=itemgetter(1,0))
python中的索引從0開始。itemgetter(1,0)按第二個元素排序，然后按第一個元素排序
慕標琳琳
TA貢獻1830條經驗 獲得超9個贊
由于您是從上到下，然后從左到右進行視覺搜索，因此此代碼要簡單得多并提供正確的結果。它基本上相當于視覺掃描，通過檢查每個“y=n”位置的所有元組，然后根據第二個數字（從左到右）對任何“y=n”元組進行排序。
為了與笛卡爾數系統更加一致，我已將圖表上的點轉換為 (x,y) 坐標，其中 X 為正（向右增加）和 y 為負（隨著它們下降而減少）。
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 loops
for 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 order
print(results)
# If you need them correlated to their original numbers
by_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" dictionary
print(by_designator_num)
或者如果你想要它更快更緊湊
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)
輸出：
[(1, -1), (3, -1), (4, -1), (2, -2), (5, -3), (2, -4), (1, -5)]
[4, 6, 3, 5, 2, 1, 7]
慕雪6442864
TA貢獻1812條經驗 獲得超5個贊
這基于對元組的第一個坐標進行排序，然后按元組的第二個坐標對代碼進行排序。即按字母順序，“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（且僅當）與 #7 關聯的元組值對在您的問題中實際上是亂序的（并且實際上應該在 #3 和 #5 之間），這將起作用。
如果不是這種情況，請參閱我的其他答案。
# Make it a dictionary, with the VALUETUPLES as the KEYS, and the designator as the value
d = {(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 tuples
l = [ (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 second
new = sorted(l, key=lambda x: x[0])
# Create a new dictionary, basically for temp sorting
new_d = {}
# This iterates through the first sorted list "new"
# and creates a dictionary where the key is the first number of value tuples
count = 0
# The extended range is because we don't know if any of the Tuple Values share any same numbers
for 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 VALUES
final_list = []
# Go through the same rage as above
for 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 order
print(final_list)
# If you need them correlated to their original numbers
by_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" dictionary
print(by_designator_num)
輸出：
[(1, 1), (1, 3), (1, 4), (1, 5), (2, 2), (3, 5), (4, 2)]
[4, 6, 3, 7, 5, 2, 1]
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