基於第一個字符，前提一個字符以輸入行

Question

對於csv文件中第一列通常爲空的數據集，我想以編程方式在每行不以雙精度開頭的行中插入額外的逗號引號。 從下面的示例中，您將看到只有某些行以"開頭，這些行是我想忽略的行。每隔一行應該額外獲得,。

"1967 Universe",1967,"1,141.0",650.6,73.0,417.5,222.6,119.6,309.8,176.0,390.8,225.0,217.8,130.0 
,1968,"1,353.4",694.7,84.3,574.4,234.8,119.2,350.1,182.1,477.3,233.9,291.1,159.5 
,1969,"1,322.3",624.8,85.2,612.4,215.8,104.7,317.0,149.9,470.5,215.6,319.0,154.6 
,1970,"1,351.5",646.8,88.1,616.7,218.3,93.1,287.4,148.1,502.9,246.5,342.9,159.1 
,1971,"1,924.6",906.1,132.9,885.7,303.6,127.3,421.1,208.1,725.4,338.6,474.6,232.1 
"1972 Universe",1972,"2,218.9","1,033.1",148.6,"1,037.2",333.3,147.3,440.8,230.1,905.4,391.5,539.3,264.3 
,1973,"1,819.5",882.1,117.0,820.5,271.9,141.6,361.4,197.5,763.2,323.9,423.1,219.0 
,1974,"1,074.4",643.8,64.4,366.2,165.4,97.2,241.3,154.5,390.1,223.7,277.6,168.4 
,1975,939.2,675.5,63.8,199.8,129.5,92.6,241.5,169.3,292.7,222.8,275.5,190.8 
,1976,"1,296.2",893.6,93.1,309.5,152.4,111.2,326.1,219.4,401.7,292.6,416.0,270.5 
,1977,"1,690.0","1,126.1",121.3,442.7,181.9,126.8,402.4,269.3,561.1,370.1,544.6,359.9 
"1978 Universe",1978,"1,800.5","1,182.6",130.6,487.3,194.4,132.9,388.0,260.8,667.6,439.3,550.5,349.6 
,1979,"1,551.8",981.5,125.4,444.8,166.9,110.2,289.1,182.4,628.0,392.1,467.7,296.7 
,1980,"1,190.6",710.4,114.5,365.7,117.9,75.7,192.0,107.9,561.9,333.0,318.9,193.7 
,1981,985.5,564.3,101.8,319.4,109.8,65.7,133.3,78.3,491.1,270.7,251.3,149.5 
,1982,"1,000.5",546.4,88.3,365.8,106.7,65.7,126.3,67.6,543.5,281.1,224.1,132.0 
,1983,"1,605.2",901.5,133.7,570.1,164.1,112.3,187.8,117.4,862.9,443.0,390.4,228.6 

我認爲這可能與sed做，但我還沒有與正則表達式熟悉，知道如何做一個有條件的更換。

我發現的唯一的例子是類似的

sed -ri 's/[^,]+/REPLACEMENT/' file.csv 

所以它可能看起來像

sed -ri 's/[^,]+/,/' file.csv 

，但我不認爲這是正確的

sed 's/\,/\,,/' 

幾乎可以工作，但它會在逗號後面的每個第一個實例之後替換逗號，而這並不是我所追求的。例如：

"2004 Universe",,2004,"2,070.1","1,613.4",90.4,366.2,197.0,131.8,370.5,295.5,960.8,756.1,541.9,430.0 
,,2005,"2,155.3","1,682.0",84.0,389.3,203.8,126.6,353.9,278.7,"1,039.0",826.8,558.6,450.0 

有人可以幫忙嗎？

Answer 1

擴展正則表達式選項-r將插入上開始逗號線的附加逗號。
circumflex表示行的開始。

Answer 2

^{有關問題的積極配方 - 前面加上,到與,開始的所有行 - 看karakfa's answer。}

要前置,所有線是不"啓動，請嘗試以下操作：

sed -i -r 's/^[^"]/,&/' file.csv 

• ^[^"]在開始捕捉任何單個字符不是一個"（[^"]） （^）並將其替換爲,，後跟匹配正則表達式（&）的字符串，即該行的第一個字符。 （在您的示例輸入中也是,）。

  • 注^的可能混淆的不同用途：作爲一方面的開口定界符/之後啓動的線錨，並且作爲反運算符作爲括號表達式中的第一個字符（字符設置）（[^...]）。

  • 需要注意的是非標準選項-r激活用於ERES支持（擴展正則表達式），而sed默認預計BREs裏面（基本正則表達式）。
    上面的作品同樣沒有-r，但是如果使用ERE是一個選項，那麼它們是可取的，因爲它們提供了更多的功能，它們的語法更像其他語言中的正則表達式。

• 不匹配，即，做開始與"任何線路，通過按原樣傳遞的任何行。

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至於你試過：

sed -ri 's/[^,]+/,/' file.csv 

比賽一個或多個（+）字符以外,上線的任何位置，因爲括號表達式（字符設置）[^,]不是錨定到字符串的開頭（與在/之後的^）。

除此之外，您的替換字符串不納入匹配字符串，所以它會與,取代的比賽，而不是前面加上它。

sed 's/\,/\,,/' 

又缺乏錨定在字符串開頭的正則表達式，因此將取代第一,上線的任何位置。 （此外，在這種情況下不需要\）。不需要

sed -i 's/^,/,&/' file 

Answer 3

FILE.CSV：

"1967 Universe",1967,"1,141.0",650.6,73.0,417.5,222.6,119.6,309.8,176.0,390.8,225.0,217.8,130.0 
,1968,"1,353.4",694.7,84.3,574.4,234.8,119.2,350.1,182.1,477.3,233.9,291.1,159.5 
,1969,"1,322.3",624.8,85.2,612.4,215.8,104.7,317.0,149.9,470.5,215.6,319.0,154.6 
,1970,"1,351.5",646.8,88.1,616.7,218.3,93.1,287.4,148.1,502.9,246.5,342.9,159.1 
,1971,"1,924.6",906.1,132.9,885.7,303.6,127.3,421.1,208.1,725.4,338.6,474.6,232.1 
"1972 Universe",1972,"2,218.9","1,033.1",148.6,"1,037.2",333.3,147.3,440.8,230.1,905.4,391.5,539.3,264.3 
,1973,"1,819.5",882.1,117.0,820.5,271.9,141.6,361.4,197.5,763.2,323.9,423.1,219.0 

用途：

$ sed 's/^,/,,/' file.csv 

輸出：

"1967 Universe",1967,"1,141.0",650.6,73.0,417.5,222.6,119.6,309.8,176.0,390.8,225.0,217.8,130.0 
,,1968,"1,353.4",694.7,84.3,574.4,234.8,119.2,350.1,182.1,477.3,233.9,291.1,159.5 
,,1969,"1,322.3",624.8,85.2,612.4,215.8,104.7,317.0,149.9,470.5,215.6,319.0,154.6 
,,1970,"1,351.5",646.8,88.1,616.7,218.3,93.1,287.4,148.1,502.9,246.5,342.9,159.1 
,,1971,"1,924.6",906.1,132.9,885.7,303.6,127.3,421.1,208.1,725.4,338.6,474.6,232.1 
"1972 Universe",1972,"2,218.9","1,033.1",148.6,"1,037.2",333.3,147.3,440.8,230.1,905.4,391.5,539.3,264.3 
,,1973,"1,819.5",882.1,117.0,820.5,271.9,141.6,361.4,197.5,763.2,323.9,423.1,219.0