def number_spoken_casual(cls, n):
"""
Split the number in groups of 2. Pronounce each one separately
but say "oh 5" for the final group if it's less than 10.
>>> EnglishUS.number_spoken_casual(302)
"three oh two"
>>> EnglishUS.number_spoken_casual(1267)
"twelve sixty seven"
"""
if n > 9999 or n % 1000 == 0:
return cls.number_spoken_full(n)
if n == 0:
return 'zero'
groups = split_num(n, 2)
if len(groups) > 1 and groups[-1] == 0:
s = "{} hundred".format(cls.lt_1000_spoken_full(groups[0]))
elif len(groups) > 1 and groups[-1] < 10:
s = "{} oh {}".format(
cls.lt_1000_spoken_full(groups[0]),
cls.lt_1000_spoken_full(groups[1])
)
else:
s = " ".join([cls.lt_1000_spoken_full(g) for g in groups])
return s
def number_spoken_full(cls, n):
"""
The basic algorithm is to split the number into groups
of 3. Pronounce each group of 3 and then append the exponent
part.
So 9999 gets split into [9, 999]. The first group is
"nine thousand". The second group is "nine hundred ninety nine"
"""
if n > MAX_NUM:
raise ValueError(
"{} is too large. Largest number is {}".format(n, MAX_NUM)
)
if n == 0:
return 'zero'
parts = []
for g, exp in zip(reversed(split_num(n, 3)), _exponents):
s = cls.lt_1000_spoken_full(g)
if s:
if exp:
s += " " + exp
parts.append(s)
return ' '.join(reversed(parts))
def number_spoken_casual(cls, n):
"""
Split the number in groups of 2. Pronounce each one separately
but say "oh 5" for the final group if it's less than 10.
>>> EnglishUS.number_spoken_casual(302)
"three oh two"
>>> EnglishUS.number_spoken_casual(1267)
"twelve sixty seven"
"""
if n > 9999 or n % 1000 == 0:
return cls.number_spoken_full(n)
if n == 0:
return 'zero'
groups = split_num(n, 2)
if len(groups) > 1 and groups[-1] == 0:
s = "{} hundred".format(cls.lt_1000_spoken_full(groups[0]))
elif len(groups) > 1 and groups[-1] < 10:
s = "{} oh {}".format(cls.lt_1000_spoken_full(groups[0]),
cls.lt_1000_spoken_full(groups[1]))
else:
s = " ".join([cls.lt_1000_spoken_full(g) for g in groups])
return s
def number_spoken_full(cls, n):
"""
The basic algorithm is to split the number into groups
of 3. Pronounce each group of 3 and then append the exponent
part.
So 9999 gets split into [9, 999]. The first group is
"nine thousand". The second group is "nine hundred ninety nine"
"""
if n > MAX_NUM:
raise ValueError("{} is too large. Largest number is {}".format(
n, MAX_NUM))
if n == 0:
return 'zero'
parts = []
for g, exp in zip(reversed(split_num(n, 3)), _exponents):
s = cls.lt_1000_spoken_full(g)
if s:
if exp:
s += " " + exp
parts.append(s)
return ' '.join(reversed(parts))
def test_split():
assert split_num(1234, 2) == [12, 34]
assert split_num(123, 2) == [1, 23]
assert split_num(123, 4) == [123]
assert split_num(1234567, 4) == [123, 4567]
def test_split():
assert split_num(1234, 2) == [12, 34]
assert split_num(123, 2) == [1, 23]
assert split_num(123, 4) == [123]
assert split_num(1234567, 4) == [123, 4567]