刪除CSV文件中帶有bash中列條件的行

Question

我有一個很大的CSV文件（5Go）。 標頭是：

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個字段）中保留包含“ 500”的行。

• 我試圖在sqlite中導入此CSV ...但是刪除崩潰...
• R也崩潰（即使從data.table讀取）

有人使用sed ， awk或其他任何命令提供解決方案嗎？

Answer 1

awk似乎要走的路：

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

NR == 1是為了保留第一行（標題），此后僅保留第13個字段為"500" 。