如何在 =OR() 语句中简化此类语句？
How can I simplify statements like these in an =OR() statement?
isnumber(search("-tr",right(j2,3
))),isnumber(search("-trus",right(j2,5))),isnumber(search(" ll",right(j2,3))),isnumber(search(" homes",right(j2,6))),isnumber(search("the ",left(j2,4))),isnumber(search(" hoa",right(j2,4))),isnumber(search("b ch",right(j2,4))),isnumber(search(" ch",right(j2,3))),isnumber(search("-trs",right(j2,4))),isnumber(search(" prop",right(j2,5))),isnumber(search(" st",right(j2,3))),isnumber(search(" av",right(j2,3))),isnumber(search(" ave",right(j2,4))),isnumber(search(" servi",right(j2,6))),isnumber(search(" maint",right(j2,6))),isnumber(search(" home",right(j2,5))),isnumber(search(" tr",right(j2,3))),isnumber(search(" assn",right(j2,5))),isnumber(search(" co",right(j2,3))),isnumber(search(" trus",right(j2,5))),isnumber(search(" trs",right(j2,4))),isnumber(search("-trs",right(j2,4))),isnumber(search(" tru",right(j2,4))),isnumber(search("jtrs",right(j2,4))),isnumber(search(" est of",right(j2,7))),isnumber(search(" trs",right(j2,4))),isnumber(value(LEFT(j2,1))),isnumber(search(" apts",right(j2,5))),isnumber(value(right(j2,3))),isnumber(search(" grp",right(j2,4))),isnumber(value(left(right(j2,4),1))),isnumber(search(" mgmt",right(j2,5))),isnumber(search(" props",right(j2,6))),isnumber(search(" tr",right(j2,3))),isnumber(search(" dev",right(j2,4))),isnumber(search(" tr",right(j2,3))),isnumber(search(" fdn",right(j2,4))),isnumber(search(" ent",right(j2,4))),isnumber(search(" PRPTS",right(j2,6))),isnumber(search(" ARPTS",right(j2,6))),isnumber(search(" univ",right(j2,5)))
所以我有这个巨大的 =OR() 语句，其中包含一堆 isnumner(search() 语句，用于检查单元格中的字符串是否以这些短语结尾。这是为了识别列表中的公司名称包含人名和公司名称。我觉得必须有一种更有效的方法。将它们全部添加到一个 isnumber(search() 这种格式 {item1|item2|item3} 不起作用。
我觉得一定有更高效的方法。
根据 提供的答案，可以使用 $-符号（这意味着“正则表达式中的字符串结尾”）来匹配字符串的结尾。另一方面，匹配字符串的开头是通过在插入符号 (^) 之后提供一个模式来完成的，表示字符串的开头。
因此，如果您想添加另一个线程中提供的公式
- (LP|JT/RS)$ ：匹配字符串末尾的 LP 或 JT/RS
- ^(ABC|DEF) : 匹配字符串开头的 ABC OR DEF
这将使公式看起来像：
=REGEXMATCH(A2, "(?i)LLC|CORPORATION|COMPANY|HOLDINGS|PARTNERS|EQUITY|(LP|JT/RS)$|^(ABC|DEF)")
参考：
isnumber(search("-tr",right(j2,3
))),isnumber(search("-trus",right(j2,5))),isnumber(search(" ll",right(j2,3))),isnumber(search(" homes",right(j2,6))),isnumber(search("the ",left(j2,4))),isnumber(search(" hoa",right(j2,4))),isnumber(search("b ch",right(j2,4))),isnumber(search(" ch",right(j2,3))),isnumber(search("-trs",right(j2,4))),isnumber(search(" prop",right(j2,5))),isnumber(search(" st",right(j2,3))),isnumber(search(" av",right(j2,3))),isnumber(search(" ave",right(j2,4))),isnumber(search(" servi",right(j2,6))),isnumber(search(" maint",right(j2,6))),isnumber(search(" home",right(j2,5))),isnumber(search(" tr",right(j2,3))),isnumber(search(" assn",right(j2,5))),isnumber(search(" co",right(j2,3))),isnumber(search(" trus",right(j2,5))),isnumber(search(" trs",right(j2,4))),isnumber(search("-trs",right(j2,4))),isnumber(search(" tru",right(j2,4))),isnumber(search("jtrs",right(j2,4))),isnumber(search(" est of",right(j2,7))),isnumber(search(" trs",right(j2,4))),isnumber(value(LEFT(j2,1))),isnumber(search(" apts",right(j2,5))),isnumber(value(right(j2,3))),isnumber(search(" grp",right(j2,4))),isnumber(value(left(right(j2,4),1))),isnumber(search(" mgmt",right(j2,5))),isnumber(search(" props",right(j2,6))),isnumber(search(" tr",right(j2,3))),isnumber(search(" dev",right(j2,4))),isnumber(search(" tr",right(j2,3))),isnumber(search(" fdn",right(j2,4))),isnumber(search(" ent",right(j2,4))),isnumber(search(" PRPTS",right(j2,6))),isnumber(search(" ARPTS",right(j2,6))),isnumber(search(" univ",right(j2,5)))
所以我有这个巨大的 =OR() 语句，其中包含一堆 isnumner(search() 语句，用于检查单元格中的字符串是否以这些短语结尾。这是为了识别列表中的公司名称包含人名和公司名称。我觉得必须有一种更有效的方法。将它们全部添加到一个 isnumber(search() 这种格式 {item1|item2|item3} 不起作用。
我觉得一定有更高效的方法。
根据
因此，如果您想添加另一个线程中提供的公式
- (LP|JT/RS)$ ：匹配字符串末尾的 LP 或 JT/RS
- ^(ABC|DEF) : 匹配字符串开头的 ABC OR DEF
这将使公式看起来像：
=REGEXMATCH(A2, "(?i)LLC|CORPORATION|COMPANY|HOLDINGS|PARTNERS|EQUITY|(LP|JT/RS)$|^(ABC|DEF)")
参考：