Convert any number (up to 1e3000) to text

Question

Given an input, calculate the correct suffix and output the number in a readable format in short scale (where 1 billion is 10^9). The suffixes must go to at least 10^3000, in which the rules for calculating them can be found here, or a list can be found here.

There is no need to output the exact numbers, this is focused on getting a huge number into a readable format.

For example:

10000 = 10.0 thousand
135933445 = 135.93 million
-2 = -2.0
-2.36734603 = -2.37
'1'+'9'*3000 = 2.0 nongennovemnonagintillion

 // out of range examples
'1'+'9'*3010 = 20000000000.0 nongennovemnonagintillion
'-1'+'9'*5000 = -inf nongennovemnonagintillion

I am aware it is similar to this question, though with 10x as many numbers and not having to read the names, there's potentially a lot more room for golfing. For the record, my result is 578 characters in Python.

Rules:

• No getting things from external resources - it must all be calculated within the code.
• External modules are fine, just don't break the above rule.
• The code should work when input is either a string, integer or float (eg. you may wish to input the super large numbers as strings).
• The output must always contain a decimal place.
• The output must be rounded if above 2 decimal places. Using the inbuilt round function is fine, I am aware there are some floating point errors, so if the occasional '.045' doesn't round up to '.05' for example, don't worry about it.
• Leaving multiple zeroes at the end is optional, as long as it doesn't go above 2 decimals (1.00 or 1.0 are both fine) and is consistent for all inputs (an input of 1 and 1.0 should output the same result).
• An input too large shouldn't cause an error, inf is a valid value since it's part of float.

Scoring:

• Score is the length of the code, including indents.
• Lowest score wins.
• Output can be either printed or returned.
• Setting the input number does not count towards length.

As a starting point, here is an ungolfed version of some code in Python to generate the list of suffixes. Feel free to build upon this or start from scratch.

a = ['', 'un','duo','tre','quattor','quin','sex','septen','octo','novem']
c = ['tillion', 'decillion', 'vigintillion', 'trigintillion', 'quadragintillion', 'quinquagintillion', 'sexagintillion', 'septuagintillion', 'octogintillion', 'nonagintillion']
d = ['', 'cen', 'duocen', 'trecen', 'quadringen', 'quingen', 'sescen', 'septingen', 'octingen', 'nongen']

num_dict = ['']
num_dict.append('thousand')
num_dict.append('million')
num_dict.append('billion')
num_dict.append('trillion')
num_dict.append('quadrillion')
num_dict.append('quintillion')
num_dict.append('sextillion')
num_dict.append('septillion')
num_dict.append('octillion')
num_dict.append('nonillion')

for prefix_hundreds in d:

    #tillion can't be used first time round
    if not prefix_hundreds:
        b = c[1:]
    else:
        b = c

    for prefix_tens in b:
        for prefix in a:
            num_dict.append(prefix_hundreds+prefix+prefix_tens)

Answer 1

To get the ball rolling then, this is my version. Only the final number is converted to a float so there is no problem with large numbers if they're within range.

In terms of speed, it takes about 0.85 seconds to calculate 1000 random integers between -1e3000 and 1e3000.

Python 2 (574)

N=str(X).split('.');K=len;E=min(1000,(K(str(abs(int(N[0]))))-1)/3);P=N[0][:(-E*3if E else K(N[0]))];L=K(P);B=''.join(N);O='%.2f'%float((P+'.'+((B[L:L+3])or'0')));m,n,o,p,q,r,s,t,u,v,w,x,y,z='illion quadr quin sex sept oct non ing ag g tre duo c t'.split();a=[''];l=a+['thousand']+[i+m for i in['m','b','tr',n,o+z,p+z,q,r,s]]+[i+j+k for i in a+[i+'en'for i in[y,x+y,w+y,n+t,o+v,'sesc',q+t,r+t,s+v]]for k in[C+m for C in[z,'dec']+[B+'int'for B in['vig','trig',n+u,o+'quag',p+u,q+'uag',r+'og',s+u]]][not i:]for j in a+['un',x,w,'quattor',o,p,q+'en',r+'o','novem']];print O,l[E]

Then with some test cases:

>>> X = 100
100.0
>>> X = '100.0'
100.0
>>> X = '1037093920'
1.04 billion
>>> X = '-1540420004036040042.400642'
-1.54 quintillion
>>> X = '1540420' + '0'*153
1.54 duoquinquagintillion
>>> X = '1540496' + '0'*1553
154.05 quingenoctodecillion

Past the limit (just to show it doesn't throw an error and still handles negative value):
>>> X = '-10935' + '6'*3005
-1093566666.67 nongennovemnonagintillion
>>> X = '10935' + '6'*3250
1.09356666667e+254 nongennovemnonagintillion
>>> X = '10935' + '6'*3500
inf nongennovemnonagintillion