我需要转换one
成1
, two
into2
等等。
有没有办法用一个库或一个类或任何东西来做到这一点?
这段代码的大部分是设置 numwords 字典,这仅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
我刚刚为 PyPI 发布了一个名为word2number的 python 模块,用于确切目的。https://github.com/akshaynagpal/w2n
安装它使用:
pip install word2number
确保您的 pip 已更新到最新版本。
用法:
from word2number import w2n
print w2n.word_to_num("two million three thousand nine hundred and eighty four")
2003984
如果有人感兴趣,我破解了一个维护字符串其余部分的版本(尽管它可能有错误,但没有对其进行过多测试)。
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
例子:
>>> 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”,则可能会出现问题。但是,这真的很粗糙。
我需要一些不同的东西,因为我的输入来自语音到文本的转换,而解决方案并不总是对数字求和。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 Andrew 的回答并对其进行了调整,以处理其他一些人们突出显示为错误的情况,并且还添加了对我上面提到的邮政编码等示例的支持。一些基本的测试用例如下所示,但我相信仍有改进的空间。
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
我需要处理一些额外的解析案例,例如序数词(“first”、“second”)、连字符(“one-hundred”)和连字符的序词(如“fifty-7”),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`
这是简单的案例方法:
>>> number = {'one':1,
... 'two':2,
... 'three':3,}
>>>
>>> number['two']
2
还是您在寻找可以处理“一万两千,一百七十二”的东西?
如果您要解析的数字数量有限,则可以轻松地将其硬编码到字典中。
对于稍微复杂的情况,您可能希望根据相对简单的数字语法自动生成此字典。与此类似的东西(当然,广义的......)
for i in range(10):
myDict[30 + i] = "thirty-" + singleDigitsDict[i]
如果您需要更广泛的东西,那么看起来您将需要自然语言处理工具。这篇文章可能是一个很好的起点。
使用 Python 包:WordToDigits
pip install wordtodigits
它可以在句子中找到以单词形式出现的数字,然后将它们转换为正确的数字格式。如果存在小数部分,还需要处理。数字的单词表示可以在文章中的任何地方。
def parse_int(string):
ONES = {'zero': 0,
'one': 1,
'two': 2,
'three': 3,
'four': 4,
'five': 5,
'six': 6,
'seven': 7,
'eight': 8,
'nine': 9,
'ten': 10,
'eleven': 11,
'twelve': 12,
'thirteen': 13,
'fourteen': 14,
'fifteen': 15,
'sixteen': 16,
'seventeen': 17,
'eighteen': 18,
'nineteen': 19,
'twenty': 20,
'thirty': 30,
'forty': 40,
'fifty': 50,
'sixty': 60,
'seventy': 70,
'eighty': 80,
'ninety': 90,
}
numbers = []
for token in string.replace('-', ' ').split(' '):
if token in ONES:
numbers.append(ONES[token])
elif token == 'hundred':
numbers[-1] *= 100
elif token == 'thousand':
numbers = [x * 1000 for x in numbers]
elif token == 'million':
numbers = [x * 1000000 for x in numbers]
return sum(numbers)
用 700 个 1 到 100 万范围内的随机数进行测试效果很好。
进行了更改,以便 text2int(scale) 将返回正确的转换。例如,text2int("hundred") => 100。
import re
numwords = {}
def text2int(textnum):
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",
'quadrillion', 'quintillion', 'sexillion', 'septillion',
'octillion', 'nonillion', 'decillion' ]
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', '')]
current = result = 0
tokens = re.split(r"[\s-]+", textnum)
for word in tokens:
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]
if scale > 1:
current = max(1, current)
current = current * scale + increment
if scale > 100:
result += current
current = 0
return result + current
Marc Burns的红宝石可以做到这一点。我最近分叉了它以增加多年的支持。您可以从 python 调用ruby 代码。
require 'numbers_in_words'
require 'numbers_in_words/duck_punch'
nums = ["fifteen sixteen", "eighty five sixteen", "nineteen ninety six",
"one hundred and seventy nine", "thirteen hundred", "nine thousand two hundred and ninety seven"]
nums.each {|n| p n; p n.in_numbers}
结果:
"fifteen sixteen"
1516
"eighty five sixteen"
8516
"nineteen ninety six"
1996
"one hundred and seventy nine"
179
"thirteen hundred"
1300
"nine thousand two hundred and ninety seven"
9297
一个快速的解决方案是使用inflect.py生成字典进行翻译。
inflect.py 有一个number_to_words()
函数,它将一个数字(例如2
)转换为它的单词形式(例如'two'
)。不幸的是,没有提供它的反向(这将允许您避免使用翻译词典路线)。同样,您可以使用该功能来构建翻译词典:
>>> import inflect
>>> p = inflect.engine()
>>> word_to_number_mapping = {}
>>>
>>> for i in range(1, 100):
... word_form = p.number_to_words(i) # 1 -> 'one'
... word_to_number_mapping[word_form] = i
...
>>> print word_to_number_mapping['one']
1
>>> print word_to_number_mapping['eleven']
11
>>> print word_to_number_mapping['forty-three']
43
如果您愿意花一些时间,可能会检查 inflect.py 函数的内部工作原理number_to_words()
并构建您自己的代码以动态执行此操作(我没有尝试过这样做)。
我采用了@recursive 的逻辑并转换为 Ruby。我还对查找表进行了硬编码,因此它并不那么酷,但可能会帮助新手了解正在发生的事情。
WORDNUMS = {"zero"=> [1,0], "one"=> [1,1], "two"=> [1,2], "three"=> [1,3],
"four"=> [1,4], "five"=> [1,5], "six"=> [1,6], "seven"=> [1,7],
"eight"=> [1,8], "nine"=> [1,9], "ten"=> [1,10],
"eleven"=> [1,11], "twelve"=> [1,12], "thirteen"=> [1,13],
"fourteen"=> [1,14], "fifteen"=> [1,15], "sixteen"=> [1,16],
"seventeen"=> [1,17], "eighteen"=> [1,18], "nineteen"=> [1,19],
"twenty"=> [1,20], "thirty" => [1,30], "forty" => [1,40],
"fifty" => [1,50], "sixty" => [1,60], "seventy" => [1,70],
"eighty" => [1,80], "ninety" => [1,90],
"hundred" => [100,0], "thousand" => [1000,0],
"million" => [1000000, 0]}
def text_2_int(string)
numberWords = string.gsub('-', ' ').split(/ /) - %w{and}
current = result = 0
numberWords.each do |word|
scale, increment = WORDNUMS[word]
current = current * scale + increment
if scale > 100
result += current
current = 0
end
end
return result + current
end
我想处理像这样的字符串two thousand one hundred and forty-six
这处理印度风格的单词中的数字,一些分数,数字和单词的组合以及加法。
def words_to_number(words):
numbers = {"zero":0, "a":1, "half":0.5, "quarter":0.25, "one":1,"two":2,
"three":3, "four":4,"five":5,"six":6,"seven":7,"eight":8,
"nine":9, "ten":10,"eleven":11,"twelve":12, "thirteen":13,
"fourteen":14, "fifteen":15,"sixteen":16,"seventeen":17,
"eighteen":18,"nineteen":19, "twenty":20,"thirty":30, "forty":40,
"fifty":50,"sixty":60,"seventy":70, "eighty":80,"ninety":90}
groups = {"hundred":100, "thousand":1_000,
"lac":1_00_000, "lakh":1_00_000,
"million":1_000_000, "crore":10**7,
"billion":10**9, "trillion":10**12}
split_at = ["and", "plus"]
n = 0
skip = False
words_array = words.split(" ")
for i, word in enumerate(words_array):
if not skip:
if word in groups:
n*= groups[word]
elif word in numbers:
n += numbers[word]
elif word in split_at:
skip = True
remaining = ' '.join(words_array[i+1:])
n+=words_to_number(remaining)
else:
try:
n += float(word)
except ValueError as e:
raise ValueError(f"Invalid word {word}") from e
return n
测试:
print(words_to_number("a million and one"))
>> 1000001
print(words_to_number("one crore and one"))
>> 1000,0001
print(words_to_number("0.5 million one"))
>> 500001.0
print(words_to_number("half million and one hundred"))
>> 500100.0
print(words_to_number("quarter"))
>> 0.25
print(words_to_number("one hundred plus one"))
>> 101
此代码适用于系列数据:
import pandas as pd
mylist = pd.Series(['one','two','three'])
mylist1 = []
for x in range(len(mylist)):
mylist1.append(w2n.word_to_num(mylist[x]))
print(mylist1)
此代码仅适用于99以下的数字。word to int和int to word(其余需要实现10-20行代码和简单的逻辑。这只是初学者的简单代码):
num = input("Enter the number you want to convert : ")
mydict = {'1': 'One', '2': 'Two', '3': 'Three', '4': 'Four', '5': 'Five','6': 'Six', '7': 'Seven', '8': 'Eight', '9': 'Nine', '10': 'Ten','11': 'Eleven', '12': 'Twelve', '13': 'Thirteen', '14': 'Fourteen', '15': 'Fifteen', '16': 'Sixteen', '17': 'Seventeen', '18': 'Eighteen', '19': 'Nineteen'}
mydict2 = ['', '', 'Twenty', 'Thirty', 'Fourty', 'fifty', 'sixty', 'Seventy', 'Eighty', 'Ninty']
if num.isdigit():
if(int(num) < 20):
print(" :---> " + mydict[num])
else:
var1 = int(num) % 10
var2 = int(num) / 10
print(" :---> " + mydict2[int(var2)] + mydict[str(var1)])
else:
num = num.lower()
dict_w = {'one': 1, 'two': 2, 'three': 3, 'four': 4, 'five': 5, 'six': 6, 'seven': 7, 'eight': 8, 'nine': 9, 'ten': 10, 'eleven': 11, 'twelve': 12, 'thirteen': 13, 'fourteen': 14, 'fifteen': 15, 'sixteen': 16, 'seventeen': '17', 'eighteen': '18', 'nineteen': '19'}
mydict2 = ['', '', 'twenty', 'thirty', 'fourty', 'fifty', 'sixty', 'seventy', 'eighty', 'ninty']
divide = num[num.find("ty")+2:]
if num:
if(num in dict_w.keys()):
print(" :---> " + str(dict_w[num]))
elif divide == '' :
for i in range(0, len(mydict2)-1):
if mydict2[i] == num:
print(" :---> " + str(i * 10))
else :
str3 = 0
str1 = num[num.find("ty")+2:]
str2 = num[:-len(str1)]
for i in range(0, len(mydict2)):
if mydict2[i] == str2:
str3 = i
if str2 not in mydict2:
print("----->Invalid Input<-----")
else:
try:
print(" :---> " + str((str3*10) + dict_w[str1]))
except:
print("----->Invalid Input<-----")
else:
print("----->Please Enter Input<-----")