我试图找到一种方法来对齐对象的两个列表(按其值排序),并通过比较其值,如果对象在两个列表中均对齐,则执行一个操作,否则,则执行另一个操作。例如,带有他的名字和年龄的孩子,并使用年龄值来对齐他们。
为了使示例简单,请尝试仅使用两个排序的整数值列表:
a = (1,2,3,6,7,11,13)
b = (2,3,4,6,7,9)
我想像这样对齐它们:
+----+----+----+----+----+----+----+----+----+
| 01 | 02 | 03 | xx | 06 | 07 | xx | 11 | 13 |
+----+ +----+ +----+----+----+
| xx | 02 | 03 | 04 | 06 | 07 | 09 | xx | xx |
+----+----+----+----+----+----+----+----+----+
要对对齐的值执行操作(例如,将其附加到list1):
list1 = [2,3,6,7]
然后对未对齐的值执行操作(例如,将其追加到list2):
list2 = [1,4,9,11,13]
或仅进行打印(如果对齐或不对齐):
1 = miss-aligned
2 = aligned
3 = aligned
4 = miss-aligned
6 = aligned
7 = aligned
9 = miss-aligned
11 = miss-aligned
13 = miss-aligned
我尝试这种方式,但是...
a = (1,2,3,6,7,11,13)
b = (2,3,4,6,7,9)
list1 = []
list2 = []
# find sames and list a different value
for x in a:
for y in b:
if x == y:
list1.append(x)
print(x," = aligned")
break
if y == b[-1]:
print(x," = miss-aligned")
list2.append(x)
# find list b different value
for y in b:
for x in a:
if x == y:
break
if x == a[-1]:
print(y," = miss-aligned")
list2.append(y)
print('Same:',list1)
print('Different:',list2)
...我得到了这个不希望的输出(对于4和9,未按正确的顺序进行操作):
1 = miss-aligned
2 = aligned
3 = aligned
6 = aligned
7 = aligned
11 = miss-aligned
13 = miss-aligned
4 = miss-aligned
9 = miss-aligned
Same: [2, 3, 6, 7]
Different: [1, 11, 13, 4, 9]
我尝试许多其他方式都没有成功。
*编辑:同一列表中没有重复的值
由于列表已排序,因此您可以执行以下操作:
def same_diff(a, b):
sames = []
diffs = []
i = 0
j = 0
while i < len(a) and j < len(b):
if a[i] == b[j]:
sames.append(a[i])
i += 1
j += 1
elif a[i] > b[j]:
diffs.append(b[j])
j += 1
else:
diffs.append(a[i])
i += 1
diffs += a[i:]
diffs += b[j:]
return sames, diffs
a = (1,2,3,6,7,11,13)
b = (2,3,4,6,7,9)
same_diff(a, b)
# ([2, 3, 6, 7], [1, 4, 9, 11, 13])
如果您正在使用值做某事而不是附加其余值,则可以在它们上循环:
def same_diff(a, b):
sames = []
diffs = []
i = 0
j = 0
while i < len(a) and j < len(b):
if a[i] == b[j]:
print 'aligned', a[i]
sames.append(a[i])
i += 1
j += 1
elif a[i] > b[j]:
print 'miss-aligned', b[j]
diffs.append(b[j])
j += 1
else:
print 'miss-aligned', a[i]
diffs.append(a[i])
i += 1
while i < len(a):
print 'miss-aligned', a[i]
diffs.append(a[i])
i += 1
while j < len(b):
print 'miss-aligned', b[j]
diffs.append(b[j])
j += 1
return sames, diffs
将导致:
same_diff(a, b)
miss-aligned 1
aligned 2
aligned 3
miss-aligned 4
aligned 6
aligned 7
miss-aligned 9
miss-aligned 11
miss-aligned 13
# ([2, 3, 6, 7], [1, 4, 9, 11, 13])
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