如何解决使用grepl删除数据框R中的行-方法论问题
在两种情况下，我已按如下方式使用grepl：
第一种方法：
rc_area <- rc_area[!grepl("Avg",rc_area$Year),]
第二种方法：
library(dplyr)
rc_area <- rc_area %>% filter(!(grepl("Avg",rc_area$Year)))
第一种方法执行时没有任何错误，对我来说不稳定。但是，第二种方法给出以下错误。
Error in env_bind_lazy(private$bindings,!!!set_names(promises,names_bindings)) : attempt to use zero-length variable name
我无法理解第二个问题。
这是请求中的数据。
structure(list(Year = c("1993-94","1994-95","2-Years'Avg:","1995-96","1996-97","1997-98","1998-99","1999-00","5-Years'Avg:","2000-01","2001-02","2002-03","2003-04","2004-05","2005-06","2006-07","2007-08","2008-09","2009-10","2010-11"),Punjab = c("1300.6","1338.7","1319.7","1327.8","1354.5","1409.9","1492.9","1609.4","1438.9","1627.2","1475.9","1512.3","1687.9","1754.3","1611.5","1762.4","1728.4","1723.5","1977.7","1931.5","1824.7","1766.8"),Sindh = c("702.9","598.3","650.6","642.3","701.8","689.3","704.1","690.4","685.6","540.1","461.1","488.3","551.2","543.9","516.9","593.2","598.1","594.0","733.5","707.7","645.3","361.2"),KPK = c("62.7","63.3","63.0","63.7","64.7","66.8","68.2","67.1","66.1","66.4","60.7","61.0","61.7","59.9","61.9","59.4","60.8","61.3","53.8","46.1"),c("","",""),Balochistan = c("120.9","124.3","122.6","128.0","130.1","151.3","158.4","148.5","143.2","142.9","116.5","163.6","159.8","161.5","148.9","206.4","193.9","136.2","190.1","183.3","191.2"),Pakistan = c("2187.1","2124.6","2155.9","2161.8","2251.1","2317.3","2423.6","2515.4","2333.8","2376.6","2114.2","2225.2","2460.6","2519.6","2339.2","2621.4","2581.2","2962.6","2883.1","2712.7","2365.3")),row.names = c(NA,22L),class = "data.frame")
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