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GeezerVison is showing github.com — Floating point rounding error in _guess_delimiter of csv module · Issue #111487 · python/cpython Open Labels 3.13only security fixesonly security fixes3.14bugs and security fixesbugs and security fixes3.15bugs and security fixesbugs and security fixesstdlibStandard Library Python modules in the Lib/ directoryStandard Library Python modules in the Lib/ directorytype-bugAn unexpected behavior, bug, or errorAn unexpected behavior, bug, or error DescriptionBug reportBug description:Hi! Using the csv module - when determining the delimiter in the So for the first chunk, when a ';' appears exactly e.g. 30 times on every row - the consistency is 100% - it passes straight away and ';' is recognised as the delimiter. When it appears 30 times on 9 rows and 29 times on 1 row, the consistency is 90%. The bug is as follows: the required consistency starts at 100% and is lowered by 1% for some iterations until it reaches the minimum threshold of 90%. Only due to floating point rounding errors the consistency becomes 0.8999999, therefore missing the last iteration. consistency = 1.0 # minimum consistency threshold threshold = 0.9 while len(delims) == 0 and consistency >= threshold: for k, v in modeList: if v[0] > 0 and v[1] > 0: if ((v[1]/total) >= consistency and (delimiters is None or k in delimiters)): delims[k] = v consistency -= 0.01 Image with "proof":
Whole method: def _guess_delimiter(self, data, delimiters): """ The delimiter /should/ occur the same number of times on each row. However, due to malformed data, it may not. We don't want an all or nothing approach, so we allow for small variations in this number. 1) build a table of the frequency of each character on every line. 2) build a table of frequencies of this frequency (meta-frequency?), e.g. 'x occurred 5 times in 10 rows, 6 times in 1000 rows, 7 times in 2 rows' 3) use the mode of the meta-frequency to determine the /expected/ frequency for that character 4) find out how often the character actually meets that goal 5) the character that best meets its goal is the delimiter For performance reasons, the data is evaluated in chunks, so it can try and evaluate the smallest portion of the data possible, evaluating additional chunks as necessary. """ data = list(filter(None, data.split('\n'))) ascii = [chr(c) for c in range(127)] # 7-bit ASCII # build frequency tables chunkLength = min(10, len(data)) iteration = 0 charFrequency = {} modes = {} delims = {} start, end = 0, chunkLength while start < len(data): iteration += 1 for line in data[start:end]: for char in ascii: metaFrequency = charFrequency.get(char, {}) # must count even if frequency is 0 freq = line.count(char) # value is the mode metaFrequency[freq] = metaFrequency.get(freq, 0) + 1 charFrequency[char] = metaFrequency for char in charFrequency.keys(): items = list(charFrequency[char].items()) if len(items) == 1 and items[0][0] == 0: continue # get the mode of the frequencies if len(items) > 1: modes[char] = max(items, key=lambda x: x[1]) # adjust the mode - subtract the sum of all # other frequencies items.remove(modes[char]) modes[char] = (modes[char][0], modes[char][1] - sum(item[1] for item in items)) else: modes[char] = items[0] # build a list of possible delimiters modeList = modes.items() total = float(min(chunkLength * iteration, len(data))) # (rows of consistent data) / (number of rows) = 100% consistency = 1.0 # minimum consistency threshold threshold = 0.9 while len(delims) == 0 and consistency >= threshold: for k, v in modeList: if v[0] > 0 and v[1] > 0: if ((v[1]/total) >= consistency and (delimiters is None or k in delimiters)): delims[k] = v consistency -= 0.01 if len(delims) == 1: delim = list(delims.keys())[0] skipinitialspace = (data[0].count(delim) == data[0].count("%c " % delim)) return (delim, skipinitialspace) # analyze another chunkLength lines start = end end += chunkLength if not delims: return ('', 0) # if there's more than one, fall back to a 'preferred' list if len(delims) > 1: for d in self.preferred: if d in delims.keys(): skipinitialspace = (data[0].count(d) == data[0].count("%c " % d)) return (d, skipinitialspace) # nothing else indicates a preference, pick the character that # dominates(?) items = [(v,k) for (k,v) in delims.items()] items.sort() delim = items[-1][1] skipinitialspace = (data[0].count(delim) == data[0].count("%c " % delim)) return (delim, skipinitialspace) CPython versions tested on:3.10 Operating systems tested on:macOS Linked PRsActivity
MetadataMetadataAssigneesLabels3.13only security fixesonly security fixes3.14bugs and security fixesbugs and security fixes3.15bugs and security fixesbugs and security fixesstdlibStandard Library Python modules in the Lib/ directoryStandard Library Python modules in the Lib/ directorytype-bugAn unexpected behavior, bug, or errorAn unexpected behavior, bug, or error Projects
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