找出兩個多維列表之間的差異

Question

我想找到多維列表與第一個元素之間的交集，但我找不到解決方案。

例子 ：

a = [[greg ,1.2 ,400 ,234] [top ,9.0 ,5.1 ,2300] [file ,5.7 ,2.2, 900] [stop, 1.6 ,6.7 ,200] 

b = [[hall,5.2 ,460 ,234] [line ,5.3 ,5.91 ,100] [file ,2.7 ,3.3, 6.4] [stop, 6.6 ,5.7 ,230] 

我需要的 ：

1.element 不在 a 但在 b 中，我只想與element[0]進行比較

期望輸出 = [[hall,5.2 ,460 ,234] [line ,5.3 ,5.91 ,100]]

2.element 不在 b 中，而在 a 中，我只想與element[0]進行比較

期望輸出 = [greg ,1.2 ,400 ,234]

然后將缺失列表附加到 a 和 b 反之亦然。

我有示例代碼，但不起作用。

at = map(tuple,a)
bt = map(tuple,b)

st1 = set(at)
st2 = set(bt)

s1 = st1.intersection(st2)
s2 = st2.intersection(st1)

Answer 1

您可以使用兩個字典並找到鍵的集合差異，其中鍵是每個第一個元素：

a = [["greg", 1.2, 400, 234], ["top", 9.0, 5.1, 2300], ["file", 5.7, 2.2, 900], ["stop", 1.6, 6.7, 200]]

b = [["hall", 5.2, 460, 234], ["line", 5.3, 5.91, 100], ["file", 2.7, 3.3, 6.4], ["stop", 6.6, 5.7, 230]]

d1 = {sub[0]: sub for sub in a}
d2 = {sub[0]: sub for sub in b}

print([d2[k] for k in d2.keys() - d1])
print([d1[k] for k in d1.keys() - d2])

輸出：

[['hall', 5.2, 460, 234], ['line', 5.3, 5.91, 100]]
[['top', 9.0, 5.1, 2300], ['greg', 1.2, 400, 234]]

正確的輸出為[['top', 9.0, 5.1, 2300], ['greg', 1.2, 400, 234]]不僅[['greg', 1.2, 400, 234]]按你的預期產量你的問題。

等效的 python 2 代碼需要使用viewkeys ：

print([d2[k] for k in d2.viewkeys() - d1])
print([d1[k] for k in d1.viewkeys() - d2])

Answer 2

a = [["greg" ,1.2 ,400 ,234], ["top" ,9.0 ,5.1 ,2300], ["file" ,5.7 ,2.2, 900], ["stop", 1.6 ,6.7 ,200]]
b = [["hall",5.2 ,460 ,234], ["line" ,5.3 ,5.91 ,100], ["file" ,2.7 ,3.3, 6.4], ["stop", 6.6 ,5.7 ,230]]

b_prime = list()
i=0
for key in a:
    if key[0] == b[i][0]:
        i = i + 1
        continue
    print "entry ",key, " is not in a"
    b_prime.append(key)
    i = i + 1

a_prime = list()
i=0
for key in b:
    if key[0] == a[i][0]:
        i = i + 1
        continue
    print "entry ",key, " is not in b"
    a_prime.append(key)
    i = i + 1


print " "
print a
print b
print "need to add these to b", b_prime
print "need to add these to a", a_prime

i=0
for key in a_prime:
    a.append(b[i])
    i = i + 1

i=0
for key in b_prime:
    b.append(a[i])
    i = i + 1

print " "
print a
print b