简化2d列表比较的输出
简化2d列表比较的输出
提问于 2020-08-28 21:07:12
我已经从一个tuples文件复制了大约2000行list。现在，我需要将list中的每个元素与list其余部分中的后续元素进行比较，并且只需要对每个元素进行一次，即，如果我接受第一个元素，并将其与列表中的每个元素进行比较，那么我可以在其余的比较中放弃它。
这是我的比较代码块：
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我的产出：
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).我需要帮助格式化输出，在我的输出中，我为一个tuple获得了五行输出，我需要简化它并获得一个清晰的输出。我想在一行中得到每一个tuple的输出，因为我有超过1500 tuples，它大约需要7500行。
我想要像这样的输出
Line 1: (20, 12, 40, 42, 45) (7, 5, 2, 2, 11) #Right side values are line numbers of the respective element in the tuple
Line 2: (40, 21, 40, 42, 49) (0, 10, 0, 0, 0)
Line 3: (6, 19, 22, 36, 48) (0, 0, 0, 0, 10)
Line 4: (2, 5, 20, 24, 33) (0, 0, 9, 0, 0)
Line 6: (3, 15, 29, 30, 37) (8, 8, 0, 0, 0)
Line 7: (20, 17, 30, 33, 43) (10, 0, 11, 0, 0)回答 4
Stack Overflow用户
回答已采纳
发布于 2020-08-28 22:05:05
注意，有些行与另一行有多个匹配。我把他们留在家里了。
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输出
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)Stack Overflow用户
发布于 2020-08-28 22:08:15
三个嵌套循环似乎不太有效，但对我来说这是最明显的解决方案。
对于列表中的每个元组，对元组中的每个数字进行迭代，并将其与列表其余部分的每个元组中位于相同位置的数字进行比较，直到找到匹配为止。
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}.")
breakStack Overflow用户
发布于 2020-08-28 22:10:45
我不确定我的问题是否正确，尤其是因为我认为你快到了。
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}.")页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/63640866
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