bash 如何在文件中特定字段的单词字符之间插入空格
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How to insert whitespace between characters of words in a specific field in a file
提问by saloua
I have a file containing 100000 lines like this
我有一个包含 100000 行这样的文件
1 0110100010010101
2 1000010010111001
3 1000011001111000
10 1011110000111110
123 0001000000100001
I would like to know how can I display efficiently just the second field by adding whitespaces between characters.
我想知道如何通过在字符之间添加空格来有效地仅显示第二个字段。
0 1 1 0 1 0 0 0 1 0 0 1 0 1 0 1
1 0 0 0 0 1 0 0 1 0 1 1 1 0 0 1
1 0 0 0 0 1 1 0 0 1 1 1 1 0 0 0
1 0 1 1 1 1 0 0 0 0 1 1 1 1 1 0
0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1
One solution would be to get the second column with awk and then add the whitespaces using sed. But as the file is too long I would like to avoid using pipes. Then I'm wondering if I can do that by just using awk.
一种解决方案是使用 awk 获取第二列，然后使用 sed 添加空格。但由于文件太长，我想避免使用管道。然后我想知道我是否可以通过使用 awk 来做到这一点。
Thanks in advance
提前致谢
回答by Kent
is this ok?
这个可以吗？
awk '{gsub(/./,"& ",);print }' yourFile
example
例子
kent$ echo "1 0110100010010101
2 1000010010111001
3 1000011001111000"|awk '{gsub(/./,"& ",);print }'
0 1 1 0 1 0 0 0 1 0 0 1 0 1 0 1
1 0 0 0 0 1 0 0 1 0 1 1 1 0 0 1
1 0 0 0 0 1 1 0 0 1 1 1 1 0 0 0
update
更新
more than 2 digits in 1st column won't work? I didn't get it:
第一列中超过 2 位数字不起作用？我没明白：
kent$ echo "133 0110100010010101
233 1000010010111001
333 1000011001111000"|awk '{gsub(/./,"& ",);print }'
0 1 1 0 1 0 0 0 1 0 0 1 0 1 0 1
1 0 0 0 0 1 0 0 1 0 1 1 1 0 0 1
1 0 0 0 0 1 1 0 0 1 1 1 1 0 0 0
gsub(/./,"& ", )
1 /./ match any single character
2 "& " & here means the matched string, in this case, each character
3 column 2
so it means, replace each character in 2nd column into the character itself + " ".
回答by Birei
One way using only awk:
一种方法只使用awk：
awk '{ gsub( /./, "& ", ); print ; }' infile
That yields:
这产生：
0 1 1 0 1 0 0 0 1 0 0 1 0 1 0 1
1 0 0 0 0 1 0 0 1 0 1 1 1 0 0 1
1 0 0 0 0 1 1 0 0 1 1 1 1 0 0 0
EDIT: Kentand I gave the same implementation, so, for this answer to be a bit more useful, I will add the sedone:
编辑：肯特和我给出了相同的实现，因此，为了让这个答案更有用，我将添加sed一个：
sed -e 's/^[^ ]* *//; s/./& /g' infile
回答by Janito Vaqueiro Ferreira Filho
Just adding a sed alternative:
只需添加一个 sed 替代方案：
sed -e 's/^.* *//;s/./& /g;s/ $//' file
Three comands:
三个命令：
- Remove the characters and spaces on the start of the line
- Replace everycharacter with itself followed by a space
- (Optional) Remove the trailing space at the end of the line
- 删除行首的字符和空格
- 将每个字符替换为自身后跟一个空格
- （可选）删除行尾的尾随空格
回答by choroba
sedsolution.
sed解决方案。
sed 's/.* //;s/\(.\)/ /g'
It adds an extra space at the end of each line. Add ;s/ $//to the expression to remove it.
它在每行的末尾添加了一个额外的空格。添加;s/ $//到表达式以将其删除。
回答by potong
This might work for you (GNU sed):
这可能对你有用（GNU sed）：
sed 's/^\S*\s*//;s/\B/ /g' /file
