Question

我有一个Python字符串，如下所示：

"5 pounds cauliflower,
cut into 1-inch florets (about 18 cups)
2 large leeks,
1 teaspoons salt
3 cups of milk"

我需要为关键字cup之前出现的每个数字添加1。

结果必须是：

"5 pounds cauliflower,
cut into 1-inch florets (about 19 cups)
2 large leeks,
1 teaspoons salt
4 cups of milk"

我有以下几点：

import re

p = re.compile('([0-9]+) cup')
for i in p.finditer(s):
    # do something with int(i.group(1)) + 1

我无法弄清楚如何只替换每次迭代中找到的数字。

我还有一个边缘情况，我可能需要用10代替9，所以我不能简单地得到数字的索引并用新数字替换该数字，因为新数字可能更长。

也欢迎不涉及正则表达式的解决方案。

Answer 1

您可以将函数作为替换字符串传递给sub函数。此函数接收match object作为参数。

处理收到的参数以为每个匹配创建替换字符串。

感谢@ctwheels的回答，我改进了我的初始正则表达式处理。

mystring = """
5 pounds cauliflower,
cut into 1-inch florets (about 19 cups)
2 large leeks,
1 teaspoons salt
4 cups of milk
"""

p = r'\d+(?= +cups?\b)'

newstring = re.sub(p, lambda x: str(int(x.group(0))+1), mystring)

print(newstring)

# outputs:
5 pounds cauliflower,
cut into 1-inch florets (about 20 cups)
2 large leeks,
1 teaspoons salt
5 cups of milk

处理单词复数化（由@CasimiretHippolyte提出）我们可以使用更广泛的模式，但更多涉及更换函数：

def repl(x):
    d = int(x.group(0).split()[0]) + 1
    return str(d) + ' cup' if d == 1 else str(d) + ' cups'

p = r'\d+ cups?'


mystring = """
5 pounds cauliflower,
cut into 1-inch florets (about 19 cups)
2 large leeks,
1 teaspoons salt
4 cups of milk
1 cup of butter
0 cups of sugar"""


newstring = re.sub(p, repl, mystring)
print(newstring)
# outputs
5 pounds cauliflower,
cut into 1-inch florets (about 20 cups)
2 large leeks,
1 teaspoons salt
5 cups of milk
2 cups of butter
1 cup of sugar

Answer 2

也不是正则表达式：

def tryParseInt(i):
    try:
        num = int(i)
    except:
        return (False,i)
    return (True,num)

txt = '''5 pounds cauliflower,
cut into 1-inch florets (about 18 cups)
2 large leeks, 
1 teaspoons salt 
3 cups of milk'''

txt2 =  txt.replace("\n"," \n ").split(" ") # add a space before newline to allow splitting
                                           # at spaces to keep newlines in-lined 
txt3 = ""   # result

for n in range(len(txt2)-1):
    prev, current =  txt2[n:n+2]
    if (current == "cup" or current == "cups" or current == "cups)"):
        isint, n = tryParseInt(prev)
        if isint:
            prev = str(n+1) 

        txt3 = txt3.strip() + " " + prev

    elif prev is not None:
        txt3 = txt3 + " " + prev

txt3 += " " + current

print(txt3.replace(" \n ","\n"))

也不是正则表达式（这是第一次尝试）：

txt = '''5 pounds cauliflower,
cut into 1-inch florets (about 18 cups)
2 large leeks,
1 teaspoons salt
3 cups of milk'''

def intOrNot(a):
    """splits a at spaces and returns a list of strings and ints where possible"""
    rv = []

    for n in a.split():
        try:
            rv.append(int(n))
        except: 
            rv.append(n)

    return rv



p = [x for x in txt.split("\n")]  # get rid on lines

t = [intOrNot(a) for a in p]      # sublists per line


for q in t:
    for idx in range(len(q)-1):
        num,cup = q[idx:idx+2]
        if isinstance(num,int) and "cup" in cup:    # do not add buttercup to the recipe
            q[idx]+=1  # add 1 to the number

text = ""
for o in t:    # puzzle output together again
    for i in o:
        if isinstance(i,int):      
            text += " " + str(i)
        else:
            text += " " + i
    text = text.strip() + "\n"

print (txt+"\n\n"+text)

输出：

5 pounds cauliflower,
cut into 1-inch florets (about 18 cups)
2 large leeks,
1 teaspoons salt
3 cups of milk

5 pounds cauliflower,
cut into 1-inch florets (about 19 cups)
2 large leeks,
1 teaspoons salt
4 cups of milk

Answer 3

您可以尝试这样的事情：

import re
pattern=r'cups?'
string_1="""5 pounds cauliflower,
cut into 1-inch florets (about 18 cups)
2 large leeks,
1 teaspoons salt
3 cups of milk"""

jk=string_1.splitlines()
for i in jk:
    wow=i.split()

    for l,k in enumerate(wow):
        if (re.search(pattern,k))!=None:
            wow[l-1]=int(wow[l-1])+1

    print(" ".join([str(i) for i in wow]))

输出：

5 pounds cauliflower,
cut into 1-inch florets (about 19 cups)
2 large leeks,
1 teaspoons salt
4 cups of milk

Answer 4

代码

See regex in use here

\d+(?= +cups?\b)

用法

See code in use here

import re

a = [
    "5 pounds cauliflower,",
    "cut into 1-inch florets (about 18 cups)",
    "2 large leeks,",
    "1 teaspoons salt",
    "3 cups of milk"
]

r = r"\d+(?= +cups?\b)"

def repl(m):
    return str(int(m.group(0)) + 1)

for s in a:
    print re.sub(r, repl, s)

用法2

此代码是对问题

下面的@CasimiretHippolyte评论的回应

See code in use here

import re

a = [
    "5 pounds cauliflower,",
    "cut into 1-inch florets (about 18 cups)",
    "2 large leeks,",
    "1 teaspoons salt",
    "3 cups of milk",
    "0 cups of milk",
    "1 cup of milk"
]

r = r"(\d+) +(cups?)\b"

def repl(m):
    x = int(m.group(1)) + 1
    return str(x) + " " + ("cup", "cups")[x > 1]

for s in a:
    print re.sub(r, repl, s)

结果

输入

5 pounds cauliflower,
cut into 1-inch florets (about 18 cups)
2 large leeks,
1 teaspoons salt
3 cups of milk

输出

5 pounds cauliflower,
cut into 1-inch florets (about 19 cups)
2 large leeks,
1 teaspoons salt
4 cups of milk

说明

\d+匹配任何数字一次或多次
(?= +cups?\b)确定以下内容的积极前瞻
- +匹配一个或多个空格字符
- cups?匹配cup或cups（s?使s可选）
- \b断言位置为单词边界

Answer 5

您可以尝试这种单线解决方案：

import re
s = """
5 pounds cauliflower,
cut into 1-inch florets (about 18 cups)
2 large leeks,
1 teaspoons salt
3 cups of milk
"""
new_s = re.sub('\d+(?=\s[a-zA-Z])', '{}', s).format(*[int(re.findall('^\d+', i)[0])+1 if re.findall('[a-zA-Z]+$', i)[0] == 'cups' else int(re.findall('^\d+', i)[0]) for i in re.findall('\d+\s[a-zA-Z]+', s)])
print(new_s)

输出：

5 pounds cauliflower,
cut into 1-inch florets (about 19 cups)
2 large leeks,
1 teaspoons salt
4 cups of milk

在与字符串中的某个正则表达式匹配的数字中添加一个

5 个答案:

代码

用法

用法2

结果

输入

输出

说明