Question

问题：如何从字符串中准确提取 6个连续数字

示例：

f657674
576767g
tt454656y
465767yy
x1234567
1234567x
n645856g
s-5656-54654657657-6576-46567785-4354
pof-user-201734545435454
4545665
345678
f546576767g
rtryty

应该给予

657674
576767
454656
465767
(blank value)
(blank value)
645856
(blank value)
(blank value)
(blank value)
345678
(blank value)
(blank value)

我尝试过的操作：（A1是第一个字符串的单元格）

=IFERROR(LOOKUP(10^6;MID(A1;ROW(INDIRECT("1:"&LEN(A1)-5));6)+0);"")

然后我将此公式拖到其他行，它会给出：

    657674    (right)
    576767    (right)
    454656    (right)
    465767    (right)
   (blank value)  (right) 
   (blank value)  (right)
    645856    (right)
    657457    (wrong)
    435454    (wrong)
    4545665   (wrong)  
    345678    (right)
    546576767 (wrong)
    (blank value)  (right)

版本：Excel 2016

Answer 1

您可能需要UDF而不是内置函数。这应该可以，但是可能需要调整。您的OP中的示例返回345678作为正确的返回值，但是两边都没有字母。因此，我假设您想要：一个连续的6个连续数字，或者两个连续的6个连续数字，两侧均带有字母。

将此添加到工作簿模块中，您可以像=return_numbers(A1)那样调用。您可能不一定要add the RegEx Reference到VBEditor。

Function return_numbers(ByVal cel As Range) As String
Dim strPattern As String
Dim regEx As New RegExp

strPattern = "[a-z]\d{6}[a-z]"

With regEx
    .Global = True
    .MultiLine = True
    .IgnoreCase = True
    .Pattern = strPattern
End With

Dim matches As Object
Set matches = regEx.Execute(cel)

If Len(cel) = 6 And IsNumeric(cel) Then
    return_numbers = cel.Value
    Set regEx = Nothing
    Exit Function
End If

If matches.Count <> 0 Then
    return_numbers = Mid(matches.Item(0), 2, Len(matches.Item(0)) - 2)
ElseIf matches.Count = 0 Then
    strPattern = "[a-z]{1}\d{6}$"
    regEx.Pattern = strPattern
    Set matches = regEx.Execute(cel)
    If matches.Count <> 0 Then
        return_numbers = Mid(matches.Item(0), 2, Len(matches.Item(0)) - 1)
    ElseIf matches.Count = 0 Then
        strPattern = "^\d{6}[a-z]{1}"
        regEx.Pattern = strPattern
        Set matches = regEx.Execute(cel)
        If matches.Count <> 0 Then
            return_numbers = Mid(matches.Item(0), 1, Len(matches.Item(0)) - 1)
        End If
    End If
End If

Set regEx = Nothing

End Function

_{如果您想加快速度，我想如果您切换If/else语句，它可能会运行得更快一些，因为它不会总是运行正则表达式，如果找到6个孤独数字}

编辑：这很笨拙。我确定有更好的正则表达式模式，所以请告诉我。

Answer 2

灵感来自布鲁斯，但配对到最小

Function ContainsSix(ByVal rng As Range) As String
    Dim re As RegExp
    Dim mc As MatchCollection
    Dim CellValue As Variant

    CellValue = rng.Cells(1, 1).Value2
    Set re = New RegExp
    With re
        .Pattern = "(?:\D|^)(\d{6})(?:\D|$)"
        .Global = True
        .MultiLine = True
        .IgnoreCase = True

        If .Test(CellValue) Then
            Set mc = .Execute(CellValue)
            ContainsSix = mc(0).SubMatches(0)
        End If
    End With
    Set re = Nothing
End Function

对正则表达式的描述：

匹配表达式但不捕获它。 [\ D | ^]
- 从2个替代品中选择
  - 任何不是数字的字符
  - 行或字符串的开头
一个编号的捕获组。 [\ d {6}]
- 任何数字，精确地重复6次
匹配表达式但不捕获它。 [\ D | $]
- 从2个替代品中选择
  - 任何不是数字的字符
  - 行尾或字符串

Answer 3

import RPi.GPIO as GPIO from time import sleep import random import decimal # motor stuff global Motor1 global Motor2 global Motor3 global p global q global dutyCycle global channel Motor1 = 16 # Input Pin Motor2 = 18 # Input Pin Motor3 = 22 # Enable Pin GPIO.setmode(GPIO.BOARD) GPIO.setup(Motor1,GPIO.OUT) GPIO.setup(Motor2,GPIO.OUT) GPIO.setup(Motor3,GPIO.OUT) p = GPIO.PWM(Motor1,30) q = GPIO.PWM(Motor2,30) p.start(0) q.start(0) dutyCycle = 30 # vibration sensor stuff channel = 40 GPIO.setup(channel, GPIO.IN) def get_wait_time(): waitTime = float(decimal.Decimal(random.randrange(1, 201))/100) return waitTime def do_the_spin(channel): remove_event_detect() i=0 global Motor1 global Motor2 global Motor3 global p global q global dutyCycle timeToSpin = 10 print "\n\nvibration detected! Run the motor!\n\n" while i<timeToSpin: try: print "FORWARD MOTION" q.ChangeDutyCycle(0) p.ChangeDutyCycle(dutyCycle) GPIO.output(Motor3,GPIO.HIGH) waitTime = get_wait_time() i+=waitTime sleep(waitTime) print "BACKWARD MOTION" p.ChangeDutyCycle(0) q.ChangeDutyCycle(dutyCycle) GPIO.output(Motor3,GPIO.HIGH) waitTime = get_wait_time() i+=waitTime sleep(waitTime) print "STOP" print 'i= '+str(i) GPIO.output(Motor3,GPIO.LOW) except KeyboardInterrupt: print "\nCleaning up and closing" p.stop() q.stop() GPIO.cleanup() quit() set_event_detect() def remove_event_detect(): print "remove_event_detect" GPIO.remove_event_detect(channel) def set_event_detect(): print "set_event_detect" GPIO.add_event_detect(channel, GPIO.BOTH, callback=do_the_spin, bouncetime=99) GPIO.add_event_callback(channel, do_the_spin) # assign function to GPIO PIN, Run function on change go_to_sleep() def go_to_sleep(): #GPIO.add_event_detect(channel, GPIO.BOTH, callback=do_the_spin, bouncetime=15000) # let us know when the pin goes HIGH or LOW print "\nWaiting for vibration..." sleep(150000) try: #GPIO.add_event_detect(channel, GPIO.BOTH, callback=do_the_spin, bouncetime=15000) # let us know when the pin goes HIGH or LOW #GPIO.add_event_callback(channel, do_the_spin) # assign function to GPIO PIN, Run function on change #do_the_spin(channel) set_event_detect() except: print "\nCleaning up and closing" p.stop() q.stop() GPIO.cleanup() quit()

请注意，如果您没有使用Excel的英语版本，那么上面的部分内容可能需要修改。

致谢

如何从字符串中准确提取6个连续数字

3 个答案: