我正在為我的離散結構課程做一個作業，要求我從輸入集A返回一個列表，該列表創建所有可能的三元組（順序無關緊要）。我正在使用我之前在程序中使用的一個函數，該函數創建一個對列表并附加在對中尚不存在的任何元素，刪除所有重復的三元組。結束集將所有元素返回為 None?？赡軐е逻@種情況的原因是什么？def listPairs(A): # type (list) -> list B = [] # Runs through every element in list A for x in A: y = A.index(x) + 1 # Makes a pair of every element in A after element x while y < len(A): B.append([x, A[y]]) y += 1 return Bprint str(listPairs([1, 2, 3, 4, 5])) + " || Expected [[1, 2], [1, 3], [1, 4], [1, 5], [2, 3], [2, 4], [2, 5], [3, 4], [3, 5], [4, 5]"def listTriples(A): # type (list) -> list B = listPairs(A) C = [] for y in B: for x in A: if x not in y: C.append(y.append(x)) if x in y: continue for z in C: if z in C: C.remove(z) return Cprint str(listTriples([1, 2, 3, 4, 5])) + " || Expected [[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 3, 4], [1, 3, 5], [1, 4, 5], ...]"
2 回答
慕尼黑的夜晚無繁華
TA貢獻1864條經驗 獲得超6個贊
你的第一個函數 listPairs（） 是可以的，但是第二個函數中的邏輯有一些問題。將每次迭代中的當前列表存儲在臨時變量中將有助于：
def listTriples(A):
B = listPairs(A)
C = []
for y in B:
for x in A:
if ((x not in y) and (x > y[len(y) - 1])):
temp = y.copy()
C.append(temp)
y.remove(x)
if x in y:
continue
return C
如果要打印每次迭代的結果，代碼應如下所示：
def listTriples(A):
B = listPairs(A)
C = []
for y in B:
for x in A:
if ((x not in y) and (x > y[len(y) - 1])):
print("x not in y")
print("Y: ",y)
print("X: ",x)
print("y.append(x): ", y.append(x))
print("new Y (temp):", y)
print("------")
temp = y.copy()
C.append(temp)
print("C: ", C)
y.remove(x)
print("Y:", y)
print("------\n\n")
if x in y:
continue
return C
最終結果：
x not in y
Y: [1, 2]
X: 3
y.append(x): None
new Y (temp): [1, 2, 3]
------
C: [[1, 2, 3]]
Y: [1, 2]
------
x not in y
Y: [1, 2]
X: 4
y.append(x): None
new Y (temp): [1, 2, 4]
------
C: [[1, 2, 3], [1, 2, 4]]
Y: [1, 2]
------
x not in y
Y: [1, 2]
X: 5
y.append(x): None
new Y (temp): [1, 2, 5]
------
C: [[1, 2, 3], [1, 2, 4], [1, 2, 5]]
Y: [1, 2]
------
x not in y
Y: [1, 3]
X: 4
y.append(x): None
new Y (temp): [1, 3, 4]
------
C: [[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 3, 4]]
Y: [1, 3]
------
x not in y
Y: [1, 3]
X: 5
y.append(x): None
new Y (temp): [1, 3, 5]
------
C: [[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 3, 4], [1, 3, 5]]
Y: [1, 3]
------
x not in y
Y: [1, 4]
X: 5
y.append(x): None
new Y (temp): [1, 4, 5]
------
C: [[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 3, 4], [1, 3, 5], [1, 4, 5]]
Y: [1, 4]
------
x not in y
Y: [2, 3]
X: 4
y.append(x): None
new Y (temp): [2, 3, 4]
------
C: [[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 3, 4], [1, 3, 5], [1, 4, 5], [2, 3, 4]]
Y: [2, 3]
------
x not in y
Y: [2, 3]
X: 5
y.append(x): None
new Y (temp): [2, 3, 5]
------
C: [[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 3, 4], [1, 3, 5], [1, 4, 5], [2, 3, 4], [2, 3, 5]]
Y: [2, 3]
------
x not in y
Y: [2, 4]
X: 5
y.append(x): None
new Y (temp): [2, 4, 5]
------
C: [[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 3, 4], [1, 3, 5], [1, 4, 5], [2, 3, 4], [2, 3, 5], [2, 4, 5]]
Y: [2, 4]
------
x not in y
Y: [3, 4]
X: 5
y.append(x): None
new Y (temp): [3, 4, 5]
------
C: [[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 3, 4], [1, 3, 5], [1, 4, 5], [2, 3, 4], [2, 3, 5], [2, 4, 5], [3, 4, 5]]
Y: [3, 4]
------
[[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 3, 4], [1, 3, 5], [1, 4, 5], [2, 3, 4], [2, 3, 5], [2, 4, 5], [3, 4, 5]]
|| Expected [[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 3, 4], [1, 3, 5], [1, 4, 5], ...]
偶然的你
TA貢獻1841條經驗 獲得超3個贊
問題出在 。的返回值為 none - 它通過附加值而不返回值進行修改。您將需要改用。C.append(y.append(x))y.append(x)yxy + [x]
添加回答
舉報
0/150
提交
取消