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Even better fix for 8f87118725

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(previous solution still did not work for numbers like 123.4 with precision < 3)
This commit is contained in:
Eduard Braun 2017-02-19 00:39:36 +01:00
parent 7c2e035644
commit f5a61eeeb3
3 changed files with 29 additions and 15 deletions
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20
scour/scour.py
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@ -2622,16 +2622,7 @@ def scourUnitlessLength(length, needsRendererWorkaround=False): # length is of
"""
if not isinstance(length, Decimal):
length = getcontext().create_decimal(str(length))
# remove trailing zeroes as we do not care for significance
intLength = length.to_integral_value()
if length == intLength:
length = Decimal(intLength)
else:
length = length.normalize()
# Gather the initial non-scientific notation version of the coordinate (we want to compare with it later)
initial_value = '{0:f}'.format(length)
initial_length = length
# reduce numeric precision
# plus() corresponds to the unary prefix plus operator and applies context precision and rounding
@ -2645,7 +2636,10 @@ def scourUnitlessLength(length, needsRendererWorkaround=False): # length is of
length = length.normalize()
# Gather the non-scientific notation version of the coordinate.
# Re-quantize from the initial value to prevent unnecessary loss of precision
# (e.g. 123.4 should become 123, not 120 or even 100)
nonsci = '{0:f}'.format(length)
nonsci = '{0:f}'.format(initial_length.quantize(Decimal(nonsci)))
if not needsRendererWorkaround:
if len(nonsci) > 2 and nonsci[:2] == '0.':
nonsci = nonsci[1:] # remove the 0, leave the dot
@ -2666,11 +2660,7 @@ def scourUnitlessLength(length, needsRendererWorkaround=False): # length is of
if len(sci) < len(nonsci):
return_value = sci
# Return the shortest representation (if they are equal prefer the original as it still has the full precision)
if len(return_value) < len(initial_value):
return return_value
else:
return initial_value
return return_value
def reducePrecision(element):