我已經從文件tuples中復制了大約 2000 多行內容?，F在我需要將其中的每個元素與其余部分中的后續元素進行比較，并且我只需要比較每個元素一次，即如果我采用第一個元素并與列表中的每個元素進行比較，那么我可以在其余部分中丟棄它比較。listtxtlistlist這是我的比較塊代碼：R = [(20, 12, 40, 42, 45), (40, 21, 40, 42, 49), (6, 19, 22, 36, 48), (2, 5, 20, 24, 33), (8, 12, 24, 28, 44), (3, 15, 29, 30, 37), (20, 17, 30, 33, 43), (3, 15, 16, 29, 42), (17, 18, 20, 35, 39), (20, 21, 23, 43, 48), (14, 24, 30, 40, 45)...]for lineno1, tup in enumerate(R): print("") # iterate over the current tuple for i, num in enumerate(tup): # compare every number in the tuple to the rest of the list for lineno2 in range(lineno1+1, len(R)): tup2 = R[lineno2] if num == tup2[i]: print(f"In line: {lineno1+1} {tup} No. '{num} is found in line {lineno2+1} {tup2}.") break我的輸出：line: 1 (20, 12, 40, 42, 45) No. 20 is found in line '7' (20, 17, 30, 33, 43). line: 1 (20, 12, 40, 42, 45) No. 12 is found in line '5' (8, 12, 24, 28, 44). line: 1 (20, 12, 40, 42, 45) No. 40 is found in line '2' (40, 21, 40, 42, 49). line: 1 (20, 12, 40, 42, 45) No. 42 is found in line '2' (40, 21, 40, 42, 49). line: 1 (20, 12, 40, 42, 45) No. 45 is found in line '11' (14, 24, 30, 40, 45). line: 2 (40, 21, 40, 42, 49) No. 21 is found in line '10' (20, 21, 23, 43, 48). line: 3 (6, 19, 22, 36, 48) No. 48 is found in line '10' (20, 21, 23, 43, 48). line: 4 (2, 5, 20, 24, 33) No. 20 is found in line '9' (17, 18, 20, 35, 39). line: 6 (3, 15, 29, 30, 37) No. 3 is found in line '8' (3, 15, 16, 29, 42). line: 6 (3, 15, 29, 30, 37) No. 15 is found in line '8' (3, 15, 16, 29, 42). line: 7 (20, 17, 30, 33, 43) No. 20 is found in line '10' (20, 21, 23, 43, 48). line: 7 (20, 17, 30, 33, 43) No. 30 is found in line '11' (14, 24, 30, 40, 45).我需要幫助格式化輸出，在我的輸出中，我得到了五行輸出，tuple我需要簡化它并獲得干凈的輸出。我想tuple在一行中獲得每個的輸出，因為我有超過 1500 個tuples，大約需要 7500 行。
4 回答
犯罪嫌疑人X
TA貢獻2080條經驗 獲得超4個贊
只是很多循環:) 請注意，某些行與另一行有多個匹配。我把它們留在里面了。
R = [(20, 12, 40, 42, 45),
(40, 21, 40, 42, 49),
(6, 19, 22, 36, 48),
(2, 5, 20, 24, 33),
(8, 12, 24, 28, 44),
(3, 15, 29, 30, 37),
(20, 17, 30, 33, 43),
(3, 15, 16, 29, 42),
(17, 18, 20, 35, 39),
(20, 21, 23, 43, 48),
(14, 24, 30, 40, 45)]
for i,x in enumerate(R):
for ie, e in enumerate(x):
for rw in range(i+1, len(R)):
if e == R[rw][ie]:
print(f"In line: {i+1} {x} No. '{e}' is found in line {rw+1} {R[rw]}")
break
輸出
In line: 0 (20, 12, 40, 42, 45) No. '20' is found in line 6 (20, 17, 30, 33, 43)
In line: 0 (20, 12, 40, 42, 45) No. '12' is found in line 4 (8, 12, 24, 28, 44)
In line: 0 (20, 12, 40, 42, 45) No. '40' is found in line 1 (40, 21, 40, 42, 49)
In line: 0 (20, 12, 40, 42, 45) No. '42' is found in line 1 (40, 21, 40, 42, 49)
In line: 0 (20, 12, 40, 42, 45) No. '45' is found in line 10 (14, 24, 30, 40, 45)
In line: 1 (40, 21, 40, 42, 49) No. '21' is found in line 9 (20, 21, 23, 43, 48)
In line: 2 (6, 19, 22, 36, 48) No. '48' is found in line 9 (20, 21, 23, 43, 48)
In line: 3 (2, 5, 20, 24, 33) No. '20' is found in line 8 (17, 18, 20, 35, 39)
In line: 5 (3, 15, 29, 30, 37) No. '3' is found in line 7 (3, 15, 16, 29, 42)
In line: 5 (3, 15, 29, 30, 37) No. '15' is found in line 7 (3, 15, 16, 29, 42)
In line: 6 (20, 17, 30, 33, 43) No. '20' is found in line 9 (20, 21, 23, 43, 48)
In line: 6 (20, 17, 30, 33, 43) No. '30' is found in line 10 (14, 24, 30, 40, 45)
胡說叔叔
TA貢獻1804條經驗 獲得超8個贊
我不確定我的問題是否正確，特別是因為我認為你已經快要問了。
R = [(20, 12, 40, 42, 45), (40, 21, 40, 42, 49),
(6, 19, 22, 36, 48), (2, 5, 20, 24, 33),
(8, 12, 24, 28, 44), (3, 15, 29, 30, 37),
(20, 17, 30, 33, 43), (3, 15, 16, 29, 42),
(17, 18, 20, 35, 39), (20, 21, 23, 43, 48),
(14, 24, 30, 40, 45),]
for i,x in enumerate(R):
a = set(x)
for j in range(i+1,len(R)):
y = R[j]
b = set(y)
ab = a&b
for n in ab:
print(f"In line: {i+1} {x} No. '{n}' is found in line {j+1} {y}.")
繁華開滿天機
TA貢獻1816條經驗 獲得超4個贊
三個嵌套循環似乎效率不高，但對我來說這是最明顯的解決方案。
對于列表中的每個元組，迭代元組中的每個數字，并將其與列表其余部分中每個元組中相同位置的數字進行比較，直到找到匹配項。
for lineno1, tup in enumerate(R):
# iterate over the current tuple
for i, num in enumerate(tup):
# compare every number in the tuple to the rest of the list
for lineno2 in range(lineno1+1, len(R)):
tup2 = R[lineno2]
if num == tup2[i]:
print(f"In line: {lineno1+1} {tup} No. '{num} is found in line {lineno2+1} {tup2}.")
break
蕪湖不蕪
TA貢獻1796條經驗 獲得超7個贊
如果我正確理解了這個問題，你只需要迭代一堆......
R = [(20, 12, 40, 42, 45), (40, 21, 40, 42, 49),
(6, 19, 22, 36, 48), (2, 5, 20, 24, 33),
(8, 12, 24, 28, 44), (3, 15, 29, 30, 37),
(20, 17, 30, 33, 43), (3, 15, 16, 29, 42),
(17, 18, 20, 35, 39), (20, 21, 23, 43, 48),
(14, 24, 30, 40, 45)...]
for i in range(len(R)):
list_a = R[i]
for d in range(len(list_a)):
val_to_find = list_a[d]
for j in range((i+1), len(R)):
list_b = R[j]
for k in range(list_b):
cur_val = list_b[k]
if(val_to_find == cur_val):
print("In line: " + string(i) + " " + string(list_a) + " No. " + string(list_a[d]) + " is found in line " + string(k) + " " + string(list_b)")
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