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Only use number representation with reduced precision if it is shorter than the initial representation.

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Before it could happen that "123" was replaced with "1e3" if precision was set to 1 which is obviously not desirable.
This commit is contained in:
Eduard Braun 2017-02-18 19:01:26 +01:00
parent b00b374e64
commit 8f87118725
1 changed files with 17 additions and 4 deletions
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21
scour/scour.py
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@ -2623,6 +2623,16 @@ 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)
# reduce numeric precision
# plus() corresponds to the unary prefix plus operator and applies context precision and rounding
length = scouringContext.plus(length)
@ -2641,6 +2651,7 @@ def scourUnitlessLength(length, needsRendererWorkaround=False): # length is of
nonsci = nonsci[1:] # remove the 0, leave the dot
elif len(nonsci) > 3 and nonsci[:3] == '-0.':
nonsci = '-' + nonsci[2:] # remove the 0, leave the minus and dot
return_value = nonsci
# Gather the scientific notation version of the coordinate which
# can only be shorter if the length of the number is at least 4 characters (e.g. 1000 = 1e3).
@ -2653,11 +2664,13 @@ def scourUnitlessLength(length, needsRendererWorkaround=False): # length is of
sci = six.text_type(length) + 'e' + six.text_type(exponent)
if len(sci) < len(nonsci):
return sci
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 nonsci
else:
return nonsci
return initial_value
def reducePrecision(element):