TA貢獻(xiàn)1826條經(jīng)驗(yàn) 獲得超6個(gè)贊
嘗試這個(gè)：
sorted(dicts,key=itemgetter(1,0))
python中的索引從0開始。itemgetter(1,0)按第二個(gè)元素排序，然后按第一個(gè)元素排序
TA貢獻(xiàn)1830條經(jīng)驗(yàn) 獲得超9個(gè)贊
由于您是從上到下，然后從左到右進(jìn)行視覺搜索，因此此代碼要簡單得多并提供正確的結(jié)果。它基本上相當(dāng)于視覺掃描，通過檢查每個(gè)“y=n”位置的所有元組，然后根據(jù)第二個(gè)數(shù)字（從左到右）對(duì)任何“y=n”元組進(jìn)行排序。
為了與笛卡爾數(shù)系統(tǒng)更加一致，我已將圖表上的點(diǎn)轉(zhuǎn)換為 (x,y) 坐標(biāo)，其中 X 為正（向右增加）和 y 為負(fù)（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]
TA貢獻(xiàn)1812條經(jīng)驗(yàn) 獲得超5個(gè)贊
這基于對(duì)元組的第一個(gè)坐標(biāo)進(jìn)行排序，然后按元組的第二個(gè)坐標(biāo)對(duì)代碼進(jìn)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(dāng)）與 #7 關(guān)聯(lián)的元組值對(duì)在您的問題中實(shí)際上是亂序的（并且實(shí)際上應(yīng)該在 #3 和 #5 之間），這將起作用。
如果不是這種情況，請(qǐng)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]
舉報(bào)