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Python: Dictionaries with variable default

+2 votes

Is it possible to have a default dictionary where the default is dependent on the key?

I was hoping something like this might work:

>>> m = defaultdict(lambda key: key+1)

But it obviously doesn't:

>>> m[3]

Traceback (most recent call last):

 File "", line 1, in 
TypeError: () takes exactly 1 argument (0 given)
posted Nov 3, 2014 by anonymous

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Using 1/3 as an example,

 >>> 1./3
 >>> print "%.50f" % (1./3)
 >>> print "%.50f" % (10./3)
 >>> print "%.50f" % (100./3)

which seems to mean real (at least default) decimal precision is limited to "double", 16 digit precision (with rounding error). Is there a way to increase the real precision, preferably as the default?
For instance, UBasic uses a "Words for fractionals", f, "Point(f)" system, where Point(f) sets the decimal display precision, .1^int(ln(65536^73)/ln(10)), with the last few digits usually garbage.
Using "90*(pi/180)*180/pi" as an example to highlight the rounding error (4 = UBasic's f default value):

 Point(2)=.1^09: 89.999999306
 Point(3)=.1^14: 89.9999999999944
 Point(4)=.1^19: 89.9999999999999998772
 Point(5)=.1^24: 89.999999999999999999999217
 Point(7)=.1^33: 89.999999999999999999999999999999823
 Point(10)=.1^48: 89.999999999999999999999999999999999999999999997686
 Point(11)=.1^52: 89.9999999999999999999999999999999999999999999999999632

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