Question

我有字符串列表，需要找到遵循此逻辑的所有内容

twodigits.twodigits  
Example| Correct?  
----------- | ------  
03.38 | Yes  
10.59  | Yes  
10.23 m3/h | No  
50.8 mm | No

我只想找“。”但是没有用......

for i in elements:
    id = i.Id
    if i.Text[2:3] == '.':
        i.Text= ''
        atext1.append(i.Text)

我发现了这个，但不确定如何在我的代码中使用它？

str_extract_all(text, "\\d{2}.\\d{2}") %>%

我目前的答案代码：

for i in elements:
    id = i.Id
    match = search('\d{2}\.\d{2}$',i.strip())
    if match: 
        atext1.append(match.string)
        i.Text= ''
        #atext1.append(i.Text)
    else:
        atext2.append(i.Text)
    atext.append(i.Text)

另一次尝试但仍然无效

for i in elements:
    id = i.Id
    k = i.Text
    atext3.append(k)

    match=search('\d{2}\.\d{2}$',k.strip())
    if match: atext4.append(match.string)
        i.Text= ''

这一个也没有成功

for i in elements:
    id = i.Id
    k = i.Text
    atext3.append(k)
    if k.split('.')[0:2].isdigit():
        i.Text= ''
        atext3.append(i.Text)

我最终得到了这个。谢谢大家的帮助

import clr
clr.AddReference('ProtoGeometry')
from Autodesk.DesignScript.Geometry import *
clr.AddReference("RevitNodes")
import Revit
clr.ImportExtensions(Revit.Elements)
clr.AddReference("RevitServices")
import RevitServices
from RevitServices.Persistence import DocumentManager
from RevitServices.Transactions import TransactionManager
from System.Collections.Generic import *
clr.AddReference("RevitAPI")
import Autodesk
from Autodesk.Revit.DB import *
from re import search

doc = DocumentManager.Instance.CurrentDBDocument
uiapp = DocumentManager.Instance.CurrentUIApplication
app = uiapp.Application


#The inputs to this node will be stored as a list in the IN variable.

#from re import search

atext3 = []
atextTrue = []
atextFalse = []
atextAll = []

#unwrap all elements to use with API
elements = []
for i in IN[0]:
    elements.append(UnwrapElement(i))

TransactionManager.Instance.EnsureInTransaction(doc)

for i in elements:
    atext3 = i.Text
    match = search('\d{2}\.\d{2}$', atext3.strip())
    if match:
        i.Text = 'deletedTEXT'      
        atextTrue.append(i.Text)
    else:
        atextFalse.append(i.Text)

    atextAll.append(i)

TransactionManager.Instance.TransactionTaskDone()

#Assign your output to the OUT variable
OUT =  atextTrue, atextFalse, atextAll

Answer 1

def fits_pattern(s):
    s = s.strip()
    return len(s) == 5 and s[2] == '.' and (s[:2] + s[-2:]).isdigit()


for test_string in ['03.38', '10.59 ', '10.23 m3/h', '50.8 mm',
                    'ab.cd', '12.3d', '1234', '12345', '', '     12.34     ']:
    print(repr(test_string))
    print(fits_pattern(test_string))

输出：

'03.38'
True
'10.59 '
True
'10.23 m3/h'
False
'50.8 mm'
False
'ab.cd'
False
'12.3d'
False
'1234'
False
'12345'
False
''
False
'     12.34     '
True

正则表达式的解决方案是：

import re

def fits_pattern(s):
    return re.match(r'\s*\d\d\.\d\d\s*', s)

\s*：0个或更多空格（包括换行符，制表符等） \d：一个数字 \.：文字.，因为.本身就是不同的东西。 r'...'：原始字符串文字，对正则表达式非常有用，可避免在字符串文字中加载\个字符。

Answer 2

from re import search

l=['03.38', '10.59 ', '10.23 m3/h', '50.8 mm']
atext1=[]
for i in l:
    match=search('\d{2}\.\d{2}$',i.strip())
    if match: atext1.append(match.string)



print(atext1) #['03.38', '10.59']

\d{2}\.\d{2}$会将l中结束的任何元素与两位数，一个句点和两位数匹配。 i.strip()删除了前导空格和尾随空格 - '10.59 '将成为'10.59'并匹配此正则表达式。

Answer 3

for i in elements:
    if not len(list(i.Text)) > 5:
        if len(i.split('.')) > 0:
           if i.split('.')[0:2].isdigit():
               if i.split('.')[3:].isdigit():
                   print("Yes")

Answer 4

# being that you provided no details on how this info is stored, other than "a list of strings", I'm assuming it's stored like so:
lst_info = ['11.11','1.11','22.11','22.22']

def is_number(s):
    try:
        float(s)
        return True
    except ValueError:
        return False


def function(p):
    new_lst = []
    piece = []
    for e in p:
        if len(e) > 4:
            if (is_number(e[0:2]), is_number(e[3:]), e[2] == '.'):
                piece.append([str(e)])
                piece.append([': Yes'])
                new_lst.append(piece)
                piece = []
        else:
            piece.append([str(e)])
            piece.append([': No'])
            new_lst.append(piece)
            piece = []
    result = ''
    for e in new_lst:
        result += (' ' * (9 - len(str(e[0])))) + (str(e[0]) + ' | ' + str(e[1]) + '\n')
    return result

print function(lst_info)
# prints
['11.11'] | [': Yes']
 ['1.11'] | [': No']
['22.11'] | [': Yes']
['22.22'] | [': Yes']

如何从2digits“。”列表项中提取仅在Iron Python中的2个数字？

4 个答案: