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_find_width_index and _handle_long_word change #2

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Merged
xi merged 2 commits into from
Nov 11, 2021
Merged

_find_width_index and _handle_long_word change #2

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e5d6d88
_find_width_index and _handle_long_word change
tiptenbrink Nov 9, 2021
7b32d0b
Apply changes, ensure min 1 char on line
tiptenbrink Nov 10, 2021
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26 changes: 26 additions & 0 deletions Lib/test/test_textwrap.py
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Original file line number Diff line number Diff line change
Expand Up @@ -1118,5 +1118,31 @@ def test_shorten_placeholder(self):
text_len=self.text_len)


class CustomWidthTestCase(BaseTestCase):
def text_len(self, text):
lengths = {
'A': 4,
'B': 2,
'Q': 0,
}

return sum(
lengths[c] if c in lengths else 1
for c in text
)

def test_zero_width_text_len(self):
text = "0QQ1234QQ56789"
self.check_wrap(text, 6, ["0QQ1234QQ5", "6789"], text_len=self.text_len)

def test_char_longer_than_width(self):
text = "AA0123"
self.check_wrap(text, 3, ["A", "A", "012", "3"], text_len=self.text_len)

def test_next_char_overflow(self):
text = "BB0123"
self.check_wrap(text, 3, ["B", "B0", "123"], text_len=self.text_len)


if __name__ == '__main__':
unittest.main()
23 changes: 21 additions & 2 deletions Lib/textwrap.py
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Original file line number Diff line number Diff line change
Expand Up @@ -199,6 +199,25 @@ def _fix_sentence_endings(self, chunks):
else:
i += 1

def _find_width_index(self, text, width):
"""_find_length_index(text : string, width : int)

Find at which index the text has the required width, since when using a
different text_len, this index will not be equal to the required width.
"""
# When using default len as self.text_len, the required index and width
# will be equal, this prevents calculation time.
if self.text_len(text[:width]) == width:
# For character widths greater than one, width can be more than the
# number of characters
return min(width, len(text))
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@xi xi Nov 10, 2021 •
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I am not sure if this heuristic works. What would happen in the following case:

lengths = {
    'A': 2,
    'B': 0,
}

def text_len(s):
    return sum(lengths.get(c, 1) for c in s)

wrap('AAABBB', 3)

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@xi xi Nov 10, 2021

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If performance is an issue, a binary search might be an option.

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@tiptenbrink tiptenbrink Nov 10, 2021

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In that case the width of text[:width] (=['AAA']) is not equal to width (=3), so it will calculate it per character. However, if you'd used len as your text_len, you'd get that the width of ['AAA'] = 3, which is correct to it skips the calculation.

Binary search might work, but it'd require disallowing negative character lengths (which I agree don't make sense but the results of binary search might make the results more different than what you'd expect).

cur_text = ''
for i, c in enumerate(text):
cur_text += c
cur_width = self.text_len(cur_text)
if cur_width > width:
return max(i, 1)

def _handle_long_word(self, reversed_chunks, cur_line, cur_len, width):
"""_handle_long_word(chunks : [string],
cur_line : [string],
Expand All @@ -217,12 +236,12 @@ def _handle_long_word(self, reversed_chunks, cur_line, cur_len, width):
# If we're allowed to break long words, then do so: put as much
# of the next chunk onto the current line as will fit.
if self.break_long_words:
end = space_left
chunk = reversed_chunks[-1]
end = self._find_width_index(chunk, space_left)
if self.break_on_hyphens and self.text_len(chunk) > space_left:
# break after last hyphen, but only if there are
# non-hyphens before it
hyphen = chunk.rfind('-', 0, space_left)
hyphen = chunk.rfind('-', 0, end)
if hyphen > 0 and any(c != '-' for c in chunk[:hyphen]):
end = hyphen + 1
cur_line.append(chunk[:end])
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