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Add new --ignore-words option to ignore 'good' words #100
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There are style issues, see Travis output |
A new command line option -I / --ignore-words is implemented to specify one or multiple files containing 'good' words which codespell will ignore.
Codecov Report
@@ Coverage Diff @@
## master #100 +/- ##
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+ Coverage 88.92% 89.44% +0.52%
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Files 3 3
Lines 578 616 +38
Branches 81 85 +4
==========================================
+ Hits 514 551 +37
Misses 51 51
- Partials 13 14 +1
Continue to review full report at Codecov.
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I have updated the commit. Any other objections to the PR? |
Can you add a test to make sure it works? |
I have added unit tests for the |
codespell_lib/tests/test_basic.py
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assert_equal(sio[1], '') # stderr | ||
assert_true('SUMMARY' in sio[0]) | ||
assert_equal(len(sio[0].split('\n')), 7) | ||
assert_true('abandonned' in sio[0].split()[-2]) | ||
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@with_setup(reload_codespell_lib, reload_codespell_lib) | ||
def test_ignore_dictonary(): | ||
"""Test ignore dictonary functionality""" |
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typo Line 171,172 should be dictionary
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Ups, thank you.
@Eric89GXL @lucasdemarchi Ready to merge? |
LGTM. @lucasdemarchi do you want to look? |
@thdot what's the format of this file of good words? I have a list that I'd like to exclude:
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It's simply one 'good' word per line. |
Note: the whitelist word needs to be identical to the word that is in dictionary.txt discovered this after encountering: #143 (comment) @thdot perhaps the |
A new command line option
-I / --ignore-words
is implemented to specify one or multiple files containing good words which codespell will ignore.This PR is related to issue #66.