#!/usr/bin/env python # ***** BEGIN LICENSE BLOCK ***** # Version: MPL 1.1/GPL 2.0/LGPL 2.1 # # The contents of this file are subject to the Mozilla Public License # Version 1.1 (the "License"); you may not use this file except in # compliance with the License. You may obtain a copy of the License at # http://www.mozilla.org/MPL/ # # Software distributed under the License is distributed on an "AS IS" # basis, WITHOUT WARRANTY OF ANY KIND, either express or implied. See the # License for the specific language governing rights and limitations # under the License. # # The Original Code is Komodo code. # # The Initial Developer of the Original Code is ActiveState Software Inc. # Portions created by ActiveState Software Inc are Copyright (C) 2000-2007 # ActiveState Software Inc. All Rights Reserved. # # Contributor(s): # ActiveState Software Inc # # Alternatively, the contents of this file may be used under the terms of # either the GNU General Public License Version 2 or later (the "GPL"), or # the GNU Lesser General Public License Version 2.1 or later (the "LGPL"), # in which case the provisions of the GPL or the LGPL are applicable instead # of those above. If you wish to allow use of your version of this file only # under the terms of either the GPL or the LGPL, and not to allow others to # use your version of this file under the terms of the MPL, indicate your # decision by deleting the provisions above and replace them with the notice # and other provisions required by the GPL or the LGPL. If you do not delete # the provisions above, a recipient may use your version of this file under # the terms of any one of the MPL, the GPL or the LGPL. # # ***** END LICENSE BLOCK ***** r"""Basic text manipulation utilities.""" import os import sys import re from pprint import pprint import logging #---- public stuff # Recipe: text_escape (0.2) def escaped_text_from_text(text, escapes="eol"): r"""Return escaped version of text. "escapes" is either a mapping of chars in the source text to replacement text for each such char or one of a set of strings identifying a particular escape style: eol replace EOL chars with '\r' and '\n', maintain the actual EOLs though too whitespace replace EOL chars as above, tabs with '\t' and spaces with periods ('.') eol-one-line replace EOL chars with '\r' and '\n' whitespace-one-line replace EOL chars as above, tabs with '\t' and spaces with periods ('.') """ #TODO: # - Add 'c-string' style. # - Add _escaped_html_from_text() with a similar call sig. import re if isinstance(escapes, basestring): if escapes == "eol": escapes = {'\r\n': "\\r\\n\r\n", '\n': "\\n\n", '\r': "\\r\r"} elif escapes == "whitespace": escapes = {'\r\n': "\\r\\n\r\n", '\n': "\\n\n", '\r': "\\r\r", '\t': "\\t", ' ': "."} elif escapes == "eol-one-line": escapes = {'\n': "\\n", '\r': "\\r"} elif escapes == "whitespace-one-line": escapes = {'\n': "\\n", '\r': "\\r", '\t': "\\t", ' ': '.'} else: raise ValueError("unknown text escape style: %r" % escapes) # Sort longer replacements first to allow, e.g. '\r\n' to beat '\r' and # '\n'. escapes_keys = escapes.keys() try: escapes_keys.sort(key=lambda a: len(a), reverse=True) except TypeError: # Python 2.3 support: sort() takes no keyword arguments escapes_keys.sort(lambda a,b: cmp(len(a), len(b))) escapes_keys.reverse() def repl(match): val = escapes[match.group(0)] return val escaped = re.sub("(%s)" % '|'.join([re.escape(k) for k in escapes_keys]), repl, text) return escaped def one_line_summary_from_text(text, length=78, escapes={'\n':"\\n", '\r':"\\r", '\t':"\\t"}): r"""Summarize the given text with one line of the given length. "text" is the text to summarize "length" (default 78) is the max length for the summary "escapes" is a mapping of chars in the source text to replacement text for each such char. By default '\r', '\n' and '\t' are escaped with their '\'-escaped repr. """ if len(text) > length: head = text[:length-3] else: head = text escaped = escaped_text_from_text(head, escapes) if len(text) > length: summary = escaped[:length-3] + "..." else: summary = escaped return summary # Recipe: break_up_words (1.0) def break_up_words(text, max_word_length=50): """Break up words(*) in the given string so no word is longer than `max_word_length`. Here a "word" means any consecutive string of characters not separated by whitespace. @param text {str} The string in which to break up words. @param max_length {int} The max word length. Default is 50. """ import re bit_is_word = True bits = [] splitter = u"\u200b" # zero-width space if isinstance(text, str): try: text = unicode(text) except UnicodeDecodeError: splitter = " " # ASCII space for bit in re.split(r"(\s+)", text, re.UNICODE): if bit_is_word: while len(bit) > max_word_length: head, bit = bit[:max_word_length], bit[max_word_length:] bits.append(head) bits.append(splitter) bits.append(bit) else: bits.append(bit) bit_is_word = not bit_is_word return ''.join(bits) def break_up_lines(text, max_line_width=80): """Break up the text in the given string so no line is longer than `max_line_width`. Any existing line endings are left unchanged. If any containing word is longer than this line width, then it will be broken up in order to meet the line width restrictions. A "word" means any consecutive string of characters not separated by whitespace. @param text {str} The string in which to break up words. @param max_line_width {int} The maximum line width allowed. Default is 80. """ import textwrap lines = [] for line in text.split("\n"): if not line: # need to manually append, otherwise textwrap eats it lines.append(line) else: lines += textwrap.wrap(line, max_line_width) return '\n'.join(lines) #---- self-test def _test(): import doctest doctest.testmod() if __name__ == "__main__": _test()