Python:文件格式

Question

我有一个for循环，它引用字典并打印出与键关联的值。代码如下：

for i in data:
    if i in dict:
        print dict[i],

我将如何格式化输出，以便每60个字符创建一条新行？例如，将字符数沿边数计算：

0001

0061

0121 LDPQACPTVIDQLTEMLKETGIPVQVIFSSGKDVDLLPQRSNKGNATQYLQQHLAMEPSQ

Answer 1

这是一个非常棘手的格式化问题，但我认为以下代码：

import sys

class EveryN(object):
  def __init__(self, n, outs):
    self.n = n        # chars/line
    self.outs = outs  # output stream
    self.numo = 1     # next tag to write
    self.tll = 0      # tot chars on this line
  def write(self, s):
    while True:
      if self.tll == 0: # start of line: emit tag
        self.outs.write('%4.4d ' % self.numo)
        self.numo += self.n
      # wite up to N chars/line, no more
      numw = min(len(s), self.n - self.tll)
      self.outs.write(s[:numw])
      self.tll += numw
      if self.tll >= self.n:
        self.tll = 0
        self.outs.write('\n')
      s = s[numw:]
      if not s: break

if __name__ == '__main__':
  sys.stdout = EveryN(60, sys.stdout)
  for i, a in enumerate('abcdefgh'):
    print a*(5+ i*5),

演示如何做到这一点--作为主脚本(5a，10b，等等，中间有空格)为演示目的运行时的输出是：

0001 aaaaa bbbbbbbbbb ccccccccccccccc dddddddddddddddddddd eeeeee
0061 eeeeeeeeeeeeeeeeeee ffffffffffffffffffffffffffffff ggggggggg
0121 gggggggggggggggggggggggggg hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
0181 hhhhhhh

Answer 2

你好像在找textwrap

--文本缠绕模块提供了两个方便的函数: work ()和fill()，以及TextWrapper (完成所有工作的类)和一个实用函数dedent()。如果您只是包装或填充一个或两个文本字符串，那么方便函数应该足够好；否则，您应该使用TextWrapper实例来提高效率。

Answer 3

# test data
data = range(10)
the_dict = dict((i, str(i)*200) for i in range( 10 ))

# your loops as a generator
lines = ( the_dict[i] for i in data if i in the_dict )

def format( line ):
    def splitter():
        k = 0
        while True:
            r = line[k:k+60] # take a 60 char block
            if r: # if there are any chars left
                yield "%04d %s" % (k+1, r) # format them
            else:
                break
            k += 60
    return '\n'.join(splitter()) # join all the numbered blocks

for line in lines:
        print format(line)