删除CSV文件中带有bash中列条件的行

Question

我有一个很大的CSV文件（5Go）。 标头是：

run number,export,downerQ,coefUpQuality,chooseMode,demandF,nbPLots,standarDevPop,nbCitys,whatWord,priceMaxWineF,marketColor,[step],giniIndexReserve,giniIndexPatch,meanQualityTotal,meanQualityMountain,meanQualityPlain,DiffExtCentral,nbcentralPlots,meanPatchByNetwork,sum_q_viti_moutain,sum_q_viti_plaine
"3","false","0.5","0.01","false","7000","10","2","10","0","70","false","0","0","0.07083333333333335","0","0","0","0","0","0","48","0"
"4","false","0.5","0.01","false","7000","10","2","10","0","70","false","0","0","0.04285714285714286","0","0","0","0","0","0","42","0"
"2","false","0.5","0.01","false","7000","10","2","10","0","70","false","0","0","0.05348837209302328","0","0","0","0","0","0","43","0"

我只想在字段[step]（第13个字段）中保留包含“ 500”的行。

• 我试图在sqlite中导入此CSV ...但是删除崩溃...
• R也崩溃（即使从data.table读取）

有人使用sed ， awk或其他任何命令提供解决方案吗？

Answer 1

awk似乎要走的路：

awk -F, 'NR == 1 || $13 == "\"500\""' filename

NR == 1是为了保留第一行（标题），此后仅保留第13个字段为"500" 。