python - 基于匹配字符的文件分离

Question

  ATOM    856  CE ALYS A 104       0.809   0.146  26.161  0.54 29.14           C
  ATOM    857  CE BLYS A 104       0.984  -0.018  26.394  0.46 31.19           C
  ATOM    858  NZ ALYS A 104       1.988   0.923  26.662  0.54 33.17           N
  ATOM    859  NZ BLYS A 104       1.708   0.302  27.659  0.46 37.61           N
  ATOM    860  OXT LYS A 104      -0.726  -6.025  27.180  1.00 26.53           O
  ATOM    862  N   LYS B 276      17.010 -16.138   9.618  1.00 41.00           N
  ATOM    863  CA  LYS B 276      16.764 -16.524  11.005  1.00 31.05           C
  ATOM    864  C   LYS B 276      16.428 -15.306  11.884  1.00 26.93           C
  ATOM    865  O   LYS B 276      16.258 -15.447  13.090  1.00 29.67           O
  ATOM    866  CB  LYS B 276      17.863 -17.347  11.617  1.00 33.62           C

我有上面的文本文件，需要根据行中第21位的差异制作两个文本文件。我写了一个可以打印所需结果的脚本。但是如果我不知道第 21 列的字符是什么，我怎么能做这个工作。以下是我试过的脚本。假设我不知道第 21 行是“A”和“B”还是“B”和“G”或任何其他组合，需要在第 21 行的基础上进行分隔。我该怎么做？

  import sys

  for fn in sys.argv[1:]:
     f=open(fn,'r')

     while 1:
        line=f.readline()
        if not line: break
        if line[21:22] == 'B':
           chns = line[0:80]
           print chns

Answer 1

• 存储上一行第 21 个字符的先前值，然后为每个不匹配的地方添加一个换行符 (这意味着另一组相同字母) 打印基于其第 21 个字符的分组行。

• 请注意，它只会根据文件中的行顺序对匹配第 21 个字符的行进行分组，这意味着未排序的行将有多个单独的组相同的第 21 个字符。

  修改文件以显示此案例:

  ATOM    856  CE ALYS A 104       0.809   0.146  26.161  0.54 29.14           C
  ATOM    857  CE BLYS A 104       0.984  -0.018  26.394  0.46 31.19           C
  ATOM    862  N   LYS B 276      17.010 -16.138   9.618  1.00 41.00           N
  ATOM    863  CA  LYS B 276      16.764 -16.524  11.005  1.00 31.05           C
  ATOM    864  C   LYS B 276      16.428 -15.306  11.884  1.00 26.93           C
  ATOM    865  O   LYS B 276      16.258 -15.447  13.090  1.00 29.67           O
  ATOM    866  CB  LYS B 276      17.863 -17.347  11.617  1.00 33.62           C
  ATOM    858  NZ ALYS A 104       1.988   0.923  26.662  0.54 33.17           N
  ATOM    859  NZ BLYS A 104       1.708   0.302  27.659  0.46 37.61           N
  ATOM    860  OXT LYS A 104      -0.726  -6.025  27.180  1.00 26.53           O
  

  产生这种情况的代码(不对行进行排序):

  import sys
  
  for fn in sys.argv[1:]:
  
      with open(fn,'r') as file:
          prev = 0
          for line in file:
              line = line.strip()
              if line[21:22] != prev:
                  # new line separator for each group
                  print ''
              print line
              prev = line[21:22]
  

  显示这种情况的示例输出:

  ATOM    856  CE ALYS A 104       0.809   0.146  26.161  0.54 29.14           C
  ATOM    857  CE BLYS A 104       0.984  -0.018  26.394  0.46 31.19           C
  
  ATOM    862  N   LYS B 276      17.010 -16.138   9.618  1.00 41.00           N
  ATOM    863  CA  LYS B 276      16.764 -16.524  11.005  1.00 31.05           C
  ATOM    864  C   LYS B 276      16.428 -15.306  11.884  1.00 26.93           C
  ATOM    865  O   LYS B 276      16.258 -15.447  13.090  1.00 29.67           O
  ATOM    866  CB  LYS B 276      17.863 -17.347  11.617  1.00 33.62           C
  
  ATOM    858  NZ ALYS A 104       1.988   0.923  26.662  0.54 33.17           N
  ATOM    859  NZ BLYS A 104       1.708   0.302  27.659  0.46 37.61           N
  ATOM    860  OXT LYS A 104      -0.726  -6.025  27.180  1.00 26.53           O
  

• 因此，如果您希望每个相同的第 21 个字符仅一组，请将所有行放在一个列表中并排序它使用 list.sort() 就可以了。

  代码(分组前先对行进行排序):

  import sys
  
  for fn in sys.argv[1:]:
  
      with open(fn,'r') as file:
  
          lines = file.readlines()
  
          # creates a list or pairs (21st char, line) within a list
          lines = [ [line[21:22], line.strip() ] for line in lines ]
  
          # sorts lines based on key (21st char)
          lines.sort()
  
          # brings back list of lines to its original state, 
          # but the order is not reverted since it is already sorted
          lines = [ line[1] for line in lines ]
  
          prev = 0
          for line in lines:
              if line[21:22] != prev:
                  # new line separator for each group
                  print ''
              print line
              prev = line[21:22]
  

  输出到:

  ATOM    856  CE ALYS A 104       0.809   0.146  26.161  0.54 29.14           C
  ATOM    857  CE BLYS A 104       0.984  -0.018  26.394  0.46 31.19           C
  ATOM    858  NZ ALYS A 104       1.988   0.923  26.662  0.54 33.17           N
  ATOM    859  NZ BLYS A 104       1.708   0.302  27.659  0.46 37.61           N
  ATOM    860  OXT LYS A 104      -0.726  -6.025  27.180  1.00 26.53           O
  
  ATOM    862  N   LYS B 276      17.010 -16.138   9.618  1.00 41.00           N
  ATOM    863  CA  LYS B 276      16.764 -16.524  11.005  1.00 31.05           C
  ATOM    864  C   LYS B 276      16.428 -15.306  11.884  1.00 26.93           C
  ATOM    865  O   LYS B 276      16.258 -15.447  13.090  1.00 29.67           O
  ATOM    866  CB  LYS B 276      17.863 -17.347  11.617  1.00 33.62           C
  

编辑:

在不同文件中写入分组行实际上不需要检查前一行的值，因为根据第 21 个字符更改文件名会打开一个新文件，从而分隔行。但在这里，我使用了 prev，这样任何以前创建的具有相同文件名的文件都不会被附加，这可能会导致文件内容困惑或不一致。

import sys

for fn in sys.argv[1:]:
    with open(fn,'r') as file:

        lines = file.readlines()

        # creates a list or pairs (21st char, line) within a list
        lines = [ [line[21:22], line ] for line in lines ]

        # sorts lines based on key (21st char)
        lines.sort()

        # brings back list of lines to its original state, 
        # but the order is not reverted since it is already sorted
        lines = [ line[1] for line in lines ]

        filename = 'file'
        prev = 0
        for line in lines:
            if line[21:22] != prev:
                # creates a new file
                file = open(filename + line[21:22] + '.txt', 'w')
            else:
                # appends to the file
                file = open(filename + line[21:22] + '.txt', 'a')

            file.write(line)
            prev = line[21:22]

如果附加以前创建的文件不是问题，则可以简化文件写入部分。但是，它存在写入具有相同文件名的文件的风险，该文件不是由脚本创建的，也不是由脚本在早期执行/ session 期间创建的。

filename = 'file'
for line in lines:
    file = open(filename + line[21:22] + '.txt', 'a')
    file.write(line)