Question

我想要比较列表，而下面的代码执行此操作并不完全符合我想要的目标。

目前它将输出：

Lines only found in TEST_1: 
4   6034    L   LAL,LALLAL
5   4231    N   AD

Lines only found in TEST_2: 
4   6034    L   LALLAL,LAL
5   4231    N   PL
6   5231    T   PAL

Lines match in both:    
1   1231    L   LA
1   1234    L   T
2   1434    A   C
3   1634    L   T

我想要的是：

Lines only found in TEST_1: 
5   4231    N   AD

Lines only found in TEST_2: 
5   4231    N   PL
6   5231    T   PAL

Lines match in both:    
1   1231    L   LA
1   1234    L   T
2   1434    A   C
3   1634    L   T
4   6034    L   LAL,LALLAL

如何使最后一列中的数据顺序无关紧要？因此4 6034 L LAL,LALLAL匹配4 6034 L LALLAL,LAL。

示例代码：

TEST_1 = [['1', '1231', 'L', 'LA'],['1', '1234', 'L', 'T'],
    ['2', '1434', 'A', 'C'],['3', '1634', 'L', 'T'],
    ['4', '6034', 'L', 'LAL,LALLAL'],['5', '4231', 'N', 'AD']]

TEST_2 = [['1', '1231', 'L', 'LA'],['1', '1234', 'L', 'T'],
    ['2', '1434', 'A', 'C'],['3', '1634', 'L', 'T'],
    ['4', '6034', 'L', 'LALLAL,LAL'],['5', '4231', 'N', 'PL'],
    ['6', '5231', 'T', 'PAL']]

MATCH_1 = []
MATCH_2 = []
NO_MATCH_1 = []
NO_MATCH_2 = []

ENTRY = [TEST_1, TEST_2, MATCH_1, MATCH_2, NO_MATCH_1, NO_MATCH_2]

for i in range(0, 2):
    for word in ENTRY[i]:
        if word not in ENTRY[1-i]:
            ENTRY[4+i].append(word)
        else:
            ENTRY[2+i].append(word)

print 'Lines only found in TEST_1:\t'
for word in ENTRY[4]:
    print '\t'.join(word)

print '\nLines only found in TEST_2:\t'
for word in ENTRY[5]:
    print '\t'.join(word)

print '\nLines match in both:\t'
for word in ENTRY[2]:
    print '\t'.join(word)

Answer 1

鉴于

TEST_1 = [['1', '1231', 'L', 'LA'],['1', '1234', 'L', 'T'],
    ['2', '1434', 'A', 'C'],['3', '1634', 'L', 'T'],
    ['4', '6034', 'L', 'LAL,LALLAL'],['5', '4231', 'N', 'AD']]

你希望它们成为集合，这样你就可以进行集合操作（{1, 2, 3, 4} - {3, 4, 5, 6} == {1, 2}），所以要设置它们：

TEST_2 = [['1', '1231', 'L', 'LA'],['1', '1234', 'L', 'T'],
    ['2', '1434', 'A', 'C'],['3', '1634', 'L', 'T'],
    ['4', '6034', 'L', 'LALLAL,LAL'],['5', '4231', 'N', 'PL'],
    ['6', '5231', 'T', 'PAL']]

TEST_1 = {tuple(frozenset(x.split(",")) for x in t) for t in TEST_1}
TEST_2 = {tuple(frozenset(x.split(",")) for x in t) for t in TEST_2}

请注意，我将每个['4', '6034', 'L', 'LALLAL,LAL']转换为({'4'}, {'6034'}, {'L'}, {'LALLAL', 'LAL'})，因为set没有订单（{'LALLAL', 'LAL'} == {'LAL', 'LALLAL'}）。

我还使用了frozenset和tuple s因为它们是不可变的，因此可以进入集合，list和普通集不能。

然后你可以打印出来：

print("ONLY IN TEST 1:")
for thing in TEST_1 - TEST_2:
    print('\t'.join(",".join(x) for x in thing))

print()

print("ONLY IN TEST 2:")
for thing in TEST_2 - TEST_1:
    print('\t'.join(",".join(x) for x in thing))

print()

print("IN BOTH:")
for thing in TEST_1 & TEST_2:
    print('\t'.join(",".join(x) for x in thing))


#>>> ONLY IN TEST 1:
#>>> 5  4231    N   AD
#>>> 
#>>> ONLY IN TEST 2:
#>>> 5  4231    N   PL
#>>> 6  5231    T   PAL
#>>> 
#>>> IN BOTH:
#>>> 1  1231    L   LA
#>>> 3  1634    L   T
#>>> 4  6034    L   LAL,LALLAL
#>>> 1  1234    L   T
#>>> 2  1434    A   C

比较列表，给定列中的内容顺序并不重要

1 个答案: