我有一个大学项目。我想将单词拆分为503并将其转换为503。 我从文本文件中提取了字符串，但是我不怎么分割它。
我要转换为测试的句子
there is five hundred three people
我想这样分裂
there,is,five hundred three,people
并接受列表以使用字典将其转换为
there is 503 people
我搜索了很多网站，但找不到任何相关信息。我尝试了 .split（），但它会分割每个单词，因此我无法将其用于项目。
如何分割字符串中的特定单词？
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问答
w52kdm 回答：如何分割字符串中的特定单词？
它是python，因此有一个用于此的库：https://github.com/careless25/text2digits
但是，如果您不喜欢使用库，则此方法（来自库）可以完全满足您的要求：
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
您可以像这样使用它：
>>> text2int("I want fifty five hot dogs for two hundred dollars.")
I want 55 hot dogs for 200 dollars.
您可以通过以下方式安装text2digits软件包：
pip install text2digits
然后使用以下程序包处理示例：
from text2digits import text2digits
t2d = text2digits.Text2Digits()
print t2d.convert("there is five hundred three people")
输出为：
>>>
there is 503 people
您将不得不使用列出的数字列表，然后在字符串中搜索所有数字并将其替换。
就是这样
strings["one","three"...] #list of numbers represented as strings
numbers[1,2,3...] #corrasponding numbers
def replaceNumbers(string): #function to replace numbers
for x in range(len(strings)): #loop through strings
#replace string with number
string= string[:string.find(x)] + str(numbers[x]) + string[string.find(x) + len(x):]
return string
然后，您需要弄清楚如何处理成百上千的东西