### Refactor Number to String

Given a number between `0-99999`, the function `number_to_english`, return the same number pass to argument in letters.

For example:

if pass `9` with argument of number_to_english this return `nine`

The task is very simple this code is very ugly, and need to refactor it.

``````def mil(n):
num={'0':'zero','1':'one','2':'two','3':'three','4':'four','5':'five','6':'six','7':'seven','8':'eight','9':'nine'}
dec={'10':'ten','11':'eleven','12':'twelve','13':'thirteen','14':'fourteen','15':'fifteen','16':'sixteen','17':'seventeen','18':'eighteen','19':'nineteen'}
dec_com={'20':'twenty','30':'thirty','40':'forty','50':'fifty','60':'sixty','70':'seventy','80':'eighty','90':'ninety'}
mil="thousand"
th=[]
n_t=[x for x in n]
for i in n_t:
if len(n_t)==2:
if i!='1' and n_t[1]=='0':
th.append(dec_com[i+'0'])
th.append(mil)
break
elif i=='1':
th.append(dec[i+n_t[1]])
th.append(mil)
break
else:
th.append(dec_com[i+'0'])
th.append(num[n_t[1]])
th.append(mil)
break
else:
th.append(num[i])
th.append(mil)
return th
def cen(n):
num={'0':'zero','1':'one','2':'two','3':'three','4':'four','5':'five','6':'six','7':'seven','8':'eight','9':'nine'}
dec={'10':'ten','11':'eleven','12':'twelve','13':'thirteen','14':'fourteen','15':'fifteen','16':'sixteen','17':'seventeen','18':'eighteen','19':'nineteen'}
dec_com={'20':'twenty','30':'thirty','40':'forty','50':'fifty','60':'sixty','70':'seventy','80':'eighty','90':'ninety'}
cen="hundred"
c=[]
n_d=[x for x in n]
for m in n_d:
if n_d[0]!='0':
c.append(num[m])
c.append(cen)
if n_d[1]=='0' and n_d[2]=='0':
break
elif n_d[1]=='0' and n_d[2]!='0':
c.append(num[n_d[2]])
break
elif n_d[1]!='1' and n_d[2]=='0':
c.append(dec_com[n_d[1]+'0'])
break
elif n_d[1]=='1':
c.append(dec[n_d[1]+n_d[2]])
break
else:
c.append(dec_com[n_d[1]+'0'])
c.append(num[n_d[2]])
break
else:
if n_d[1]=='0' and n_d[2]=='0':
break
elif n_d[1]=='0' and n_d[2]!='0':
c.append(num[n_d[2]])
break
elif n_d[1]!='1' and n_d[2]=='0':
c.append(dec_com[n_d[1]+'0'])
break
elif n_d[1]!='1' and n_d[2]!='0':
c.append(dec_com[n_d[1]+'0'])
c.append(num[n_d[2]])
break
elif n_d[1]=='1':
c.append(dec[n_d[1]+n_d[2]])
break

return c
def number_to_english(n):
num={0:'zero',1:'one',2:'two',3:'three',4:'four',5:'five',6:'six',7:'seven',8:'eight',9:'nine'}
dec={10:'ten',11:'eleven',12:'twelve',13:'thirteen',14:'fourteen',15:'fifteen',16:'sixteen',17:'seventeen',18:'eighteen',19:'nineteen'}
dec_com={20:'twenty',30:'thirty',40:'forty',50:'fifty',60:'sixty',70:'seventy',80:'eighty',90:'ninety'}
th=[]
c=[]
m='{0:,}'.format(n)
m=m.split(",")
try:
if n<0 or type(n)==float or n>99999:
pass
elif n<10:
c.append(num[n])
elif n<20:
c.append(dec[n])
elif n%10==0 and n<99:
c.append(dec_com[n])
elif n<99:
k=list(str(n))
c.append(dec_com[int(k[0]+'0')])
c.append(num[int(k[1])])
else:
c=cen(m[1])
th=mil(m[0])
except IndexError:
if n<0 or type(n)==float or n>99999:
pass
elif n<10:
c.append(num[n])
elif n<20:
c.append(dec[n])
elif n%10==0 and n<99:
c.append(dec_com[n])
elif n<99:
k=list(str(n))
c.append(dec_com[int(k[0]+'0')])
c.append(num[int(k[1])])
else:
c=cen(m[0])
t=[]
t.extend(th)
t.extend(c)
return " ".join(t)``````