如何在python中重新排列字符串行

如何解决如何在python中重新排列字符串行

我有如下所述的字符串，并且每组包含n个行，我想按如下所述排列字符串..

real   51.85ms
sys    22.41ms
usr    29.70ms
[www.ms786.com] 2345 sunset follow
Multidd    567890KB

real   61.85ms
sys    32.41ms
usr    27.70ms
[www.ms586.com] 4345 sunset follow
Multidd    4567890KB

real   51.85ms
sys    12.41ms
usr    41.70ms
[www.ms186.com] 7345 sunset follow
Multidd    8967890KB

TO

[www.ms786.com] 2345 sunset follow
Multidd    567890KB
real   51.85ms
sys    22.41ms
usr    29.70ms
Multidd    567890KB

[www.ms586.com] 4345 sunset follow
Multidd    4567890KB
real   61.85ms
sys    32.41ms
usr    27.70ms

[www.ms186.com] 7345 sunset follow
Multidd    8967890KB
real   51.85ms
sys    12.41ms
usr    41.70ms`

您能帮忙如何在字符串本身中重新排列它吗？非常感谢您的帮助

解决方法

此解决方案可以有效地做到这一点，方法是找到组标题并将其移回到正确的位置：

s = """real   51.85ms
sys    22.41ms
usr    29.70ms
[www.ms786.com] 2345 sunset follow
Multidd    567890KB

real   61.85ms
sys    32.41ms
usr    27.70ms
[www.ms586.com] 4345 sunset follow
Multidd    4567890KB

real   51.85ms
sys    12.41ms
usr    41.70ms
[www.ms186.com] 7345 sunset follow
Multidd    8967890KB"""

def move_header_back(lines,header_prefix,base_index = 0):
    # Find group header positions
    group_header_positions = [base_index - 1] + [i + 1 for i in range(len(lines)) if lines[i].startswith(header_prefix)]
    # If no group positions found (there is one as we inserted `0` regardless)
    if 1 == len(group_header_positions):
        return lines
    # Shift back groups headers
    for i in range(1,len(group_header_positions)):
        group_header = lines.pop(group_header_positions[i] - 1)
        lines.insert(group_header_positions[i - 1] + 1,group_header)
    return lines

def fix_group_positions(lines):
    move_header_back(lines,header_prefix = '[',base_index=0)
    move_header_back(lines,header_prefix = 'Multidd',base_index=1)
    # Join back lines
    return '\n'.join(lines)

lines = [l for l in s.split('\n') if l]
print(fix_group_positions(lines))

输出：

[www.ms786.com] 2345 sunset follow
Multidd    567890KB
real   51.85ms
sys    22.41ms
usr    29.70ms
[www.ms586.com] 4345 sunset follow
Multidd    4567890KB
real   61.85ms
sys    32.41ms
usr    27.70ms
[www.ms186.com] 7345 sunset follow
Multidd    8967890KB
real   51.85ms
sys    12.41ms
usr    41.70ms

,

此代码通过定义每个块的行排列方式来工作。例如，对于"340125"，旧的第3行将变为新行0，旧的第4行->新的行1，依此类推。
只要所有块的结构都相同，它将适用于所有块结构，每块最多10行。

class BlockRearrange:
    def __init__(self,order):
        self.order = list(map(int,order))
        self.blocksize = len(self.order)
    
    def reorder(self,s):
        result = ""
        lines = s.splitlines()

        for block_pos in range(0,len(lines),self.blocksize):
            part = lines[block_pos:block_pos+self.blocksize]
            part_length = len(part)

            for n in self.order:
                if n < part_length:
                    result += part[n] + "\n"
        
            if part_length < self.blocksize:
                break

        return result

# Create instance with formula for how to rearrange the blocks.
# The order-string needs to contain every line of the block,# even those not changing positions.
br = BlockRearrange("340125")

print(br.reorder(s))