我需要转换one成1,two转入2等等.
有没有办法用库或类或任何东西来做到这一点？
这段代码的大部分是设置numwords dict,它只在第一次调用时完成.
def text2int(textnum, numwords={}):
if not numwords:
units = [
"zero", "one", "two", "three", "four", "five", "six", "seven", "eight",
"nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen",
"sixteen", "seventeen", "eighteen", "nineteen",
]
tens = ["", "", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"]
scales = ["hundred", "thousand", "million", "billion", "trillion"]
numwords["and"] = (1, 0)
for idx, word in enumerate(units): numwords[word] = (1, idx)
for idx, word in enumerate(tens): numwords[word] = (1, idx * 10)
for idx, word in enumerate(scales): numwords[word] = (10 ** (idx * 3 or 2), 0)
current = result = 0
for word in textnum.split():
if word not in numwords:
raise Exception("Illegal word: " + word)
scale, increment = numwords[word]
current = current * scale + increment
if scale > 100:
result += current
current = 0
return result + current
print text2int("seven billion one hundred million thirty one thousand three hundred thirty seven")
#7100031337
如果有人有兴趣,我会修改一个维护字符串其余部分的版本(虽然它可能有bug,但没有测试过多).
def text2int (textnum, numwords={}):
if not numwords:
units = [
"zero", "one", "two", "three", "four", "five", "six", "seven", "eight",
"nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen",
"sixteen", "seventeen", "eighteen", "nineteen",
]
tens = ["", "", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"]
scales = ["hundred", "thousand", "million", "billion", "trillion"]
numwords["and"] = (1, 0)
for idx, word in enumerate(units): numwords[word] = (1, idx)
for idx, word in enumerate(tens): numwords[word] = (1, idx * 10)
for idx, word in enumerate(scales): numwords[word] = (10 ** (idx * 3 or 2), 0)
ordinal_words = {'first':1, 'second':2, 'third':3, 'fifth':5, 'eighth':8, 'ninth':9, 'twelfth':12}
ordinal_endings = [('ieth', 'y'), ('th', '')]
textnum = textnum.replace('-', ' ')
current = result = 0
curstring = ""
onnumber = False
for word in textnum.split():
if word in ordinal_words:
scale, increment = (1, ordinal_words[word])
current = current * scale + increment
if scale > 100:
result += current
current = 0
onnumber = True
else:
for ending, replacement in ordinal_endings:
if word.endswith(ending):
word = "%s%s" % (word[:-len(ending)], replacement)
if word not in numwords:
if onnumber:
curstring += repr(result + current) + " "
curstring += word + " "
result = current = 0
onnumber = False
else:
scale, increment = numwords[word]
current = current * scale + increment
if scale > 100:
result += current
current = 0
onnumber = True
if onnumber:
curstring += repr(result + current)
return curstring
例:
>>> text2int("I want fifty five hot dogs for two hundred dollars.")
I want 55 hot dogs for 200 dollars.
如果您有"200美元",可能会出现问题.但是,这真的非常粗糙.
我刚刚向PyPI发布了一个名为word2number的python模块,用于确切的目的.https://github.com/akshaynagpal/w2n
安装使用:
pip install word2number
确保您的点子更新到最新版本.
用法:
from word2number import w2n
print w2n.word_to_num("two million three thousand nine hundred and eighty four")
2003984
感谢代码片段...给我节省了很多时间!
我需要处理一些额外的解析案例,例如序数词("第一","第二"),带连字符的词("一百")和带连字符的序数词("第五十七"),所以我添加了几行:
def text2int(textnum, numwords={}):
if not numwords:
units = [
"zero", "one", "two", "three", "four", "five", "six", "seven", "eight",
"nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen",
"sixteen", "seventeen", "eighteen", "nineteen",
]
tens = ["", "", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"]
scales = ["hundred", "thousand", "million", "billion", "trillion"]
numwords["and"] = (1, 0)
for idx, word in enumerate(units): numwords[word] = (1, idx)
for idx, word in enumerate(tens): numwords[word] = (1, idx * 10)
for idx, word in enumerate(scales): numwords[word] = (10 ** (idx * 3 or 2), 0)
ordinal_words = {'first':1, 'second':2, 'third':3, 'fifth':5, 'eighth':8, 'ninth':9, 'twelfth':12}
ordinal_endings = [('ieth', 'y'), ('th', '')]
textnum = textnum.replace('-', ' ')
current = result = 0
for word in textnum.split():
if word in ordinal_words:
scale, increment = (1, ordinal_words[word])
else:
for ending, replacement in ordinal_endings:
if word.endswith(ending):
word = "%s%s" % (word[:-len(ending)], replacement)
if word not in numwords:
raise Exception("Illegal word: " + word)
scale, increment = numwords[word]
current = current * scale + increment
if scale > 100:
result += current
current = 0
return result + current`
我需要一些不同的东西，因为我的输入是从语音到文本的转换，解决方案并不总是将数字相加。例如，“我的邮政编码是一二三四五”不应转换为“我的邮政编码是15”。
我接受了安德鲁的回答，并对其进行了调整，以处理人们强调为错误的其他一些情况，并且还增加了对示例的支持，例如我上面提到的邮政编码。下面显示了一些基本的测试用例，但我确定仍有改进的空间。
def is_number(x):
if type(x) == str:
x = x.replace(',', '')
try:
float(x)
except:
return False
return True
def text2int (textnum, numwords={}):
units = [
'zero', 'one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight',
'nine', 'ten', 'eleven', 'twelve', 'thirteen', 'fourteen', 'fifteen',
'sixteen', 'seventeen', 'eighteen', 'nineteen',
]
tens = ['', '', 'twenty', 'thirty', 'forty', 'fifty', 'sixty', 'seventy', 'eighty', 'ninety']
scales = ['hundred', 'thousand', 'million', 'billion', 'trillion']
ordinal_words = {'first':1, 'second':2, 'third':3, 'fifth':5, 'eighth':8, 'ninth':9, 'twelfth':12}
ordinal_endings = [('ieth', 'y'), ('th', '')]
if not numwords:
numwords['and'] = (1, 0)
for idx, word in enumerate(units): numwords[word] = (1, idx)
for idx, word in enumerate(tens): numwords[word] = (1, idx * 10)
for idx, word in enumerate(scales): numwords[word] = (10 ** (idx * 3 or 2), 0)
textnum = textnum.replace('-', ' ')
current = result = 0
curstring = ''
onnumber = False
lastunit = False
lastscale = False
def is_numword(x):
if is_number(x):
return True
if word in numwords:
return True
return False
def from_numword(x):
if is_number(x):
scale = 0
increment = int(x.replace(',', ''))
return scale, increment
return numwords[x]
for word in textnum.split():
if word in ordinal_words:
scale, increment = (1, ordinal_words[word])
current = current * scale + increment
if scale > 100:
result += current
current = 0
onnumber = True
lastunit = False
lastscale = False
else:
for ending, replacement in ordinal_endings:
if word.endswith(ending):
word = "%s%s" % (word[:-len(ending)], replacement)
if (not is_numword(word)) or (word == 'and' and not lastscale):
if onnumber:
# Flush the current number we are building
curstring += repr(result + current) + " "
curstring += word + " "
result = current = 0
onnumber = False
lastunit = False
lastscale = False
else:
scale, increment = from_numword(word)
onnumber = True
if lastunit and (word not in scales):
# Assume this is part of a string of individual numbers to
# be flushed, such as a zipcode "one two three four five"
curstring += repr(result + current)
result = current = 0
if scale > 1:
current = max(1, current)
current = current * scale + increment
if scale > 100:
result += current
current = 0
lastscale = False
lastunit = False
if word in scales:
lastscale = True
elif word in units:
lastunit = True
if onnumber:
curstring += repr(result + current)
return curstring
一些测试...
one two three -> 123 three forty five -> 345 three and forty five -> 3 and 45 three hundred and forty five -> 345 three hundred -> 300 twenty five hundred -> 2500 three thousand and six -> 3006 three thousand six -> 3006 nineteenth -> 19 twentieth -> 20 first -> 1 my zip is one two three four five -> my zip is 12345 nineteen ninety six -> 1996 fifty-seventh -> 57 one million -> 1000000 first hundred -> 100 I will buy the first thousand -> I will buy the 1000 # probably should leave ordinal in the string thousand -> 1000 hundred and six -> 106 1 million -> 1000000
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