Python Text Processing (2)

Access the sub-string (obtain the data of some fields in the record) solution: slice, however, you can only obtain one field [python] theline = 'I Love python' afield = theline [] print afield result Love. If you need to consider the length of the field, use struct. unpack [python] #-*-coding: gb2312-*-import struct baseformat = '1s 6x 6s '# first, baseformat defines how to group strings. x indicates skipping, s indicates taking out # So the code above indicates taking away 1 character, skipping 6 characters, and taking away 6 Characters theline = 'I Love Python So much' numremain = len (theline) -struct. calcsize (baseformat) # Of course, the length of the theline string must be greater than 6 + 6 + 1, therefore, we need to calculate the remaining number of strings format = '% s % s' % (baseformat, numremain) # regrouping. Now we include the remaining strings l, s1, s2 = struct. unpack (format, theline) # use format and theline to perform the unpack operation on print l, s1, and s2. If you want to obtain data in a group of 5 bytes, use the slicing method with list derivation [python] theline = 'I Love Python So much' fivers = [theline [k: k + 5] for k in xrange (0, len (theline), 5)] print fivers result ['I Lov', 'e pyt', 'Hon s', 'O muc ', 'H'] replacing the substring in a string solution: [python] def expand (format, d, marker = '"', safe = False): if safe: def lookup (w): return d. get (w, w. join (marker * 2) else: def lookup (w): return d [w] parts = format. split (marker) parts [1: 2] = map (lookup, parts [1: 2]) return ''. join (parts) if _ name _ = '_ main _': print expand ('just "a" test', {'A ': 'one'}) after Python2.4, you can use string. template Class to complete this task [python] #-*-coding: gb2312-*-import string # generate a Template from the string, where the identifier is $ marked new_style = string. template ('this is $ thing ') # input a dictionary parameter print new_style.substitute ({'thing': 5}) print new_style.substitute ({'thing ': 'test'}) # You can also pass the keyword parameter print new_style.substitute (thing = 5) print new_style.substitute (thing = 'test ') result this is 5 this is test this is 5 this is test in Python2.3, you can use the following method: [python] #-*-coding: gb2312-*-old_style = 'this is % (thing) s' # The identifier is placed in brackets. Use the % operator. Put the dictionary print old_style % {'the' on the right side ': 5} print old_style % {'thing ': 'test'} This method can also be used in Python2.4, but it is less concise than the new method.