如何通過將它們與LINUX中的另一個數據文件進行比較來將相同的等級賦予一個id列表？

Question

我有根據它們的值（列3）從1到600排列的id（列2）的列表。我有另外一個同樣ID的列表，但是排名不同，因爲它們的差別是不同的。我怎樣才能把file2中id的第一個secound列表的排列順序與file1中的第一個id列表相混淆？例如：

file1: 
    rank list-of-ids values 
    1 HOUSAM69708729 0.4468 
    2 HOCANM106363549 0.4434 
    3 HOCANM10845509 0.4268 
    4 HOCANM11098662 0.4203 
    5 HOUSAM68571374 0.3896 
    6 HOUSAM69990251 0.3895 
    7 HONLDM716072164 0.3893 
    8 HOUSAM69756113 0.3656 
    9 HOCANM11098658 0.3593 
    10 HOUSAM66626020 0.3538 

file2: 
list-of-ids values 
HOCANM106363549 0.4832 
HOUSAM69708729 0.4199 
HOCANM10845509 0.4143 
HOUSAM69990251 0.3887 
HOCANM11098662 0.3792 
HOUSAM69756113 0.365 
HOUSAM68571374 0.3649 
HONLDM716072164 0.3600 
HOUSAM66626020 0.3593 
HOCANM11098658 0.3545 

輸出文件應與排名從文件1來進行file2中：

output: 
rank list-of-ids values 
2 HOCANM106363549 0.4832 
1 HOUSAM69708729 0.4199 
3 HOCANM10845509 0.4143 
6 HOUSAM69990251 0.3887 
4 HOCANM11098662 0.3792 
8 HOUSAM69756113 0.365 
5 HOUSAM68571374 0.3649 
7 HONLDM716072164 0.3600 
10 HOUSAM66626020 0.3593 
9 HOCANM11098658 0.3545 

任何建議，好嗎？請注意，真實數據沒有任何標題，因此輸出不應該有標題。

Answer 1

AWK溶液：

awk 'NR==FNR{ a[$2]=$1; next }{ print a[$1],$1,$2 }' file1 file2 

• NR==FNR - 處理所述第一輸入文件（即file1）

• a[$2]=$1 - 捕獲rank值（第一場$1）到陣列a索引編號爲list-of-ids個值（第二場$2）

• next - 跳躍到下一個記錄（file1）

• print a[$1],$1,$2 - 從所述第二輸入文件file2打印字段（$1, $2）與對應rank值a[$1]

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輸出：

2 HOCANM106363549 0.4832 
1 HOUSAM69708729 0.4199 
3 HOCANM10845509 0.4143 
6 HOUSAM69990251 0.3887 
4 HOCANM11098662 0.3792 
8 HOUSAM69756113 0.365 
5 HOUSAM68571374 0.3649 
7 HONLDM716072164 0.3600 
10 HOUSAM66626020 0.3593 
9 HOCANM11098658 0.3545 

Answer 2

另一種選擇，使用'join'

$ join -1 2 -2 1 -o 1.1,2.1,2.2 <(sort -k 2 file1) <(sort -k 1 file2) 
2 HOCANM106363549 0.4832 
3 HOCANM10845509 0.4143 
9 HOCANM11098658 0.3545 
4 HOCANM11098662 0.3792 
7 HONLDM716072164 0.3600 
10 HOUSAM66626020 0.3593 
5 HOUSAM68571374 0.3649 
1 HOUSAM69708729 0.4199 
8 HOUSAM69756113 0.365                   
6 HOUSAM69990251 0.3887                   
ranks list-of-ids values 

誠然，這不處理的頭很乾淨。你已經接受的解決辦法，但我喜歡這個工具，而不是很多人都知道它;）

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編輯：如果源數據沒有任何標題，則該命令的偉大工程：

$ cat file1 
    1 HOUSAM69708729 0.4468 
    2 HOCANM106363549 0.4434                 
    3 HOCANM10845509 0.4268                 
    4 HOCANM11098662 0.4203                 
    5 HOUSAM68571374 0.3896 
    6 HOUSAM69990251 0.3895 
    7 HONLDM716072164 0.3893 
    8 HOUSAM69756113 0.3656 
    9 HOCANM11098658 0.3593 
    10 HOUSAM66626020 0.3538 
$ cat file2 
HOCANM106363549 0.4832 
HOUSAM69708729 0.4199 
HOCANM10845509 0.4143 
HOUSAM69990251 0.3887 
HOCANM11098662 0.3792 
HOUSAM69756113 0.365 
HOUSAM68571374 0.3649 
HONLDM716072164 0.3600 
HOUSAM66626020 0.3593 
HOCANM11098658 0.3545 
$ join -1 2 -2 1 -o 1.1,2.1,2.2 <(sort -k 2 file1) <(sort -k 1 file2) 
2 HOCANM106363549 0.4832 
3 HOCANM10845509 0.4143 
9 HOCANM11098658 0.3545 
4 HOCANM11098662 0.3792 
7 HONLDM716072164 0.3600 
10 HOUSAM66626020 0.3593 
5 HOUSAM68571374 0.3649 
1 HOUSAM69708729 0.4199 
8 HOUSAM69756113 0.365 
6 HOUSAM69990251 0.3887 

如果任一文件中確實包含了頭，那麼你可以只用grep出來的「排序」前：

$ cat file1 
ranks list-of-ids values 
    1 HOUSAM69708729 0.4468 
    2 HOCANM106363549 0.4434 
    3 HOCANM10845509 0.4268 
    4 HOCANM11098662 0.4203 
    5 HOUSAM68571374 0.3896 
    6 HOUSAM69990251 0.3895 
    7 HONLDM716072164 0.3893 
    8 HOUSAM69756113 0.3656 
    9 HOCANM11098658 0.3593 
    10 HOUSAM66626020 0.3538 
$ cat file2 
list-of-ids values 
HOCANM106363549 0.4832 
HOUSAM69708729 0.4199 
HOCANM10845509 0.4143 
HOUSAM69990251 0.3887 
HOCANM11098662 0.3792 
HOUSAM69756113 0.365 
HOUSAM68571374 0.3649 
HONLDM716072164 0.3600 
HOUSAM66626020 0.3593 
HOCANM11098658 0.3545 
$ join -1 2 -2 1 -o 1.1,2.1,2.2 <(grep -v "list-of-ids" file1 | sort -k 2) <(grep -v "list-of-ids" file2 | sort -k 1) 
2 HOCANM106363549 0.4832 
3 HOCANM10845509 0.4143 
9 HOCANM11098658 0.3545 
4 HOCANM11098662 0.3792 
7 HONLDM716072164 0.3600 
10 HOUSAM66626020 0.3593 
5 HOUSAM68571374 0.3649 
1 HOUSAM69708729 0.4199 
8 HOUSAM69756113 0.365 
6 HOUSAM69990251 0.3887