我需要将one转换为1，two转换为2等。
有没有使用库、类或其他方法来完成此操作的方式？
有没有一种方法将数字单词转换为整数？
104
- Llyod
5
3好的，我会尽力为您翻译。以下是需要翻译的内容：参见：https://dev59.com/LnVD5IYBdhLWcg3wJIIK - tzot
也许这个会有帮助：http://pastebin.com/WwFCjYtt - alvas
5如果有人仍在寻找答案，我已经从下面所有答案中获得灵感，并创建了一个Python包: https://github.com/careless25/text2digits - stackErr
1我已经使用以下示例来开发和扩展此过程，但是为了将来的参考，我将其翻译成了西班牙语：https://github.com/elbaulp/text2digits_es - Alejandro Alcalde
1任何不寻找Python解决方案的人，这里有一个并行的C#问题：将单词（字符串）转换为Int，这里是Java的一个：在Java中将单词转换为数字。 - Tomerikoo
19个回答
1
进行了更改，使得text2int(scale)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
- Dawa
2
我认为100的正确英文拼写是“one hundred”。 - recursive
@recursive 你说得完全正确，但这段代码的优点在于它处理了“hundredth”（也许这就是Dawa试图强调的）。从描述听起来，其他类似的代码需要“one hundredth”，而这并不总是常用术语（例如，“她挑选出第一百个要丢弃的物品”）。 - Neil
1
我正在寻找一个能帮助我支持上述所有以及更多边缘情况的库，比如序数词（第一，第二），更大的数字，运算符等等，我找到了这个numwords-to-nums。
你可以通过以下方式进行安装。
这是一个基本的例子
你可以通过以下方式进行安装。
pip install numwords_to_nums
这是一个基本的例子
from numwords_to_nums.numwords_to_nums import NumWordsToNum
num = NumWordsToNum()
result = num.numerical_words_to_numbers("twenty ten and twenty one")
print(result) # Output: 2010 and 21
eval_result = num.evaluate('Hey calculate 2+5')
print(eval_result) # Output: 7
result = num.numerical_words_to_numbers('first')
print(result) # Output: 1st
- Harshit Dalal
1
一个快速的解决方案是使用inflect.py生成翻译字典。
inflect.py有一个
如果您愿意花些时间，或许可以检查inflect.py中的`number_to_words()`函数的内部工作，并构建自己的代码以实现动态转换（我尚未尝试过这样做）。
inflect.py有一个
number_to_words()函数，可以将数字（例如2）转换为它的单词形式（例如'two'）。不幸的是，它的反向功能（可以避免使用翻译字典）没有提供。尽管如此，您仍然可以使用该函数来构建翻译字典：>>> 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
如果您愿意花些时间，或许可以检查inflect.py中的`number_to_words()`函数的内部工作，并构建自己的代码以实现动态转换（我尚未尝试过这样做）。
- alukach
0
这个程序处理印度风格的数字，一些分数，数字和文字的组合以及加法。
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
- Hemant Hegde
1
我进行了更多的测试，“seventeen hundred”等于1700，“one thousand and seven hundred”也等于1700，但是“one thousand seven hundred”则等于（一千零七）百，即1007 * 100 = 100700。在技术上说，“one thousand seven hundred”和“one thousand AND seven hundred”是否正确？！ - Hemant Hegde
0
我拿 @recursive 的 logic 并将其转换为 Ruby。 我还硬编码了查找表，因此它不太酷，但可能有助于新手理解正在发生的事情。
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
我想要处理像 two thousand one hundred and forty-six 这样的字符串。
- whatapalaver
0
这是一个很酷的解决方案，所以我采用了@recursive的Python代码，并借助ChatGPT将其转换为C#，并对其进行了简化、格式化和紧凑处理。
是的，我不得不给ChatGPT提供了大量的指令。这花费了我一些时间，但现在它已经完成了。
我相信这段代码以及算法的工作原理更清晰、更易于理解。
是的，我不得不给ChatGPT提供了大量的指令。这花费了我一些时间，但现在它已经完成了。
我相信这段代码以及算法的工作原理更清晰、更易于理解。
public class Parser
{
public static int ParseInt(string s)
{
Dictionary<string, (int scale, int increment)> numwords = new Dictionary<string, (int, int)>
{
{"and", (1, 0)}, {"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)}, {"billion", (1000000000, 0)}
};
int current = 0;
int result = 0;
foreach (string word in s.Replace("-", " ").Split())
{
var (scale, increment) = numwords[word];
current = current * scale + increment;
if (scale > 100)
{
result += current;
current = 0;
}
}
return result + current;
}
}
- cruz0e
-1
我发现了更快的方法：
Da_Unità_a_Cifre = {'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}
Da_Lettere_a_Decine = {"tw": 20, "th": 30, "fo": 40, "fi": 50, "si": 60, "se": 70, "ei": 80, "ni": 90, }
elemento = input("insert the word:")
Val_Num = 0
try:
elemento.lower()
elemento.strip()
Unità = elemento[elemento.find("ty")+2:] # è uguale alla str: five
if elemento[-1] == "y":
Val_Num = int(Da_Lettere_a_Decine[elemento[0] + elemento[1]])
print(Val_Num)
elif elemento == "onehundred":
Val_Num = 100
print(Val_Num)
else:
Cifre_Unità = int(Da_Unità_a_Cifre[Unità])
Cifre_Decine = int(Da_Lettere_a_Decine[elemento[0] + elemento[1]])
Val_Num = int(Cifre_Decine + Cifre_Unità)
print(Val_Num)
except:
print("invalid input")
- user20549697
-2
这段代码适用于一系列数据：
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)
- WireData india
1
什么是
w2n？它在任何地方都没有被定义。 - Tomerikoo-3
此代码仅适用于小于99的数字。对于其他数字的转换，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<-----")
- Shriram Jadhav
2
1请解释一下这段代码是做什么的，以及它是如何实现的。这样你的回答对那些对编程不太了解的人更有价值。 - Luuklag
如果用户输入数字，程序将返回其对应的单词，反之亦然。例如，5->five，而Five->5。该程序适用于小于100的数字，但只需添加几行代码即可扩展到任何范围。 - Shriram Jadhav
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