Python 将父前缀编号展开为给定的子编号
我有一个父前缀编号和一个以相同前缀开头的子编号。我想将父编号扩展到目标子编号 例如，以一种包括所有前缀并显示目标的方式Python 将父前缀编号展开为给定的子编号,python,Python,我有一个父前缀编号和一个以相同前缀开头的子编号。我想将父编号扩展到目标子编号 例如，以一种包括所有前缀并显示目标的方式 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ + + Example #1 + Example #2
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ +
+ Example #1 + Example #2 + Example #3 +
+ parent=123, target=1235 + parent=123, target=12354 + parent=123, target=123073 +
+ + + +
+ Expansion would be: + Expansion would be: + Expansion would be: +
+ + + +
+ 123-0 + 123-0 + 123-0-0 +
+ 123-1 + 123-1 + 123-0-1 +
+ 123-2 + 123-2 + 123-0-2 +
+ 123-4 + 123-4 + 123-0-3 +
+ 123-5 --> target + 123-5-0 + 123-0-4 +
+ 123-6 + 123-5-1 + 123-0-5 +
+ 123-7 + 123-5-2 + 123-0-6 +
+ 123-8 + 123-5-4 --> target + 123-0-7-0 +
+ 123-9 + 123-5-5 + 123-0-7-1 +
+ + 123-5-6 + 123-0-7-2 +
+ + 123-5-7 + 123-0-7-3 --> target +
+ + 123-5-8 + 123-0-7-4 +
+ + 123-5-9 + 123-0-7-5 +
+ + 123-6 + 123-0-7-6 +
+ + 123-7 + 123-0-7-7 +
+ + 123-8 + 123-0-7-8 +
+ + 123-9 + 123-0-7-9 +
+ + + 123-0-8 +
+ + + 123-0-9 +
+ + + 123-1 +
+ + + 123-2 +
+ + + 123-3 +
+ + + 123-4 +
+ + + 123-5 +
+ + + 123-6 +
+ + + 123-7 +
+ + + 123-8 +
+ + + 123-9 +
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
- 在示例#1中，目标长度为4位，并且只有一级扩展 需要
- 在示例#2中，目标为5位长度和两个级别 需要扩大
- 在示例#3中，目标是6位长度和三级扩展 需要
parent="1312314"
child="13123147674"
l = len(parent)
target=child[l:]
t=int(target[0])
for k in range(t):
print(parent + str(k))
print(parent + str(t) + " --> target")
for k in range(t+1,10):
print(parent + str(k))
13123140
13123141
13123142
13123143
13123144
13123145
13123146
13123147 --> target
13123148
13123149
Example #4
parent=12345, target1=1234538, target2=1234570924, target3=123459
Expansion would be:
123450
123451
123452
1234530
1234531
1234532
1234533
1234534
1234535
1234536
1234537
1234538 --> target
1234539
123454
123455
123456
12345700
12345701
12345702
12345703
12345704
12345705
12345706
12345707
12345708
123457090
123457091
1234570920
1234570921
1234570922
1234570923
1234570924 --> target
1234570925
1234570926
1234570927
1234570928
1234570929
123457093
123457094
123457095
123457096
123457097
123457098
123457099
1234571
1234572
1234573
1234574
1234575
1234576
1234577
1234578
1234579
123458
123459 --> target
parent='123'
child='123073'
if len(child)> len(parent):
tuple_list =[]
nums_to_append=child[len(parent):]
l = len(nums_to_append)
for i,num in enumerate(nums_to_append):
for j in range(10):
if num == str(j):
tuple_list.append((parent,j+1))
parent = parent+num
if i==(l-1):
print(parent, "--> target")
break
else:
print(parent+str(j))
for parent,num in tuple_list[::-1]:
for j in range(num,10):
print(parent+str(j))
你可以试试这个。诀窍是
result.sort（）1230
1231
1232
1233
1234
1235
12350
12351
12352
12353
12354--> target
12355
12356
12357
12358
12359
1236
1237
1238
1239
parent="12345"
target1="1234538"
target2="1234570924"
target3="123459"
targets = [target1, target2, target3]
result = []
for target in targets:
for i in range(len(parent), len(target)):
for x in range(0, 10):
y = target[0:i] + str(x)
result.append(y)
result = list(set(result))
result.sort()
for a in result:
print (a + ("--> target" if a in targets else ""))
因为您在目标[0]中仅获取1个元素。所以是7。如果需要其他元素，则需要循环target[i]。在如何循环target[i]这一部分中，我一直在研究如何使其工作。如果要使其无限大（例如，第二个数字比第一个数字长），则我认为这需要一个递归函数。幸运的是，只有当第二个数字比第一个数字长2个字符时，这才有效one@Shijith你好谢谢你的帮助。我一直在尝试，它似乎工作得很好。我想知道当有多个目标（1到n个目标）时，您是否可以帮助如何使其工作。我用示例4更新了我的原始帖子，其中有3个目标和相应的输出。谢谢你的帮助。它工作非常出色，甚至是一个更紧凑的解决方案。如果你有机会回答我原始帖子中的第二个问题，请你帮助我，那就是处理目标不止一个的情况，如例4所示。再次感谢你可以用目标列表来包装它。我更新了我的答案
1230
1231
1232
1233
1234
1235
12350
12351
12352
12353
12354--> target
12355
12356
12357
12358
12359
1236
1237
1238
1239
parent="12345"
target1="1234538"
target2="1234570924"
target3="123459"
targets = [target1, target2, target3]
result = []
for target in targets:
for i in range(len(parent), len(target)):
for x in range(0, 10):
y = target[0:i] + str(x)
result.append(y)
result = list(set(result))
result.sort()
for a in result:
print (a + ("--> target" if a in targets else ""))