使用 grep 过滤大于和小于范围的值 [英] Filter with grep a value greater and less than range
问题描述
我有一个包含这些数据的 csv:
I have a csv with this data:
0,M,19,finnish,english swedish german
9,M,30,urdu,english
122,F,26,finnish,english swedish german
83,M,20,finnish,english french swedish
44,F,20,finnish,english swedish
10,F,29,finnish,english
而且我需要一个使用 GREP 的过滤器,而不是采用大于 10 且小于 99 的用户值(第一列).
And I need a filter using GREP than take user value (first column) great than 10 and less than 99.
这是我最好的动作:
cat demographic_info.csv | grep -e "1[0-9]*"
推荐答案
假设您希望只匹配 10
到 99
的数字(即 11
到 98
包含),你可以使用
Assuming you want to match numbers from 10
to 99
exclusively (that is, 11
to 98
inclusively), you can use
grep -E '^(1[1-9]|[2-8][0-9]|9[0-8]),' file
数字范围模式在如何用正则表达式匹配X和Y之间的数字?,我只需要删除 ?:
因为 POSIX ERE 不支持非捕获组.
The numeric range pattern is automatically generated at How to match numbers between X and Y with regexp?, I just needed to remove ?:
as POSIX ERE does not support non-capturing groups.
然而,
awk -F, '$1 < 99 && $1 > 10' file
看起来更适合这项任务.它使用逗号作为字段分隔符,并检查第一个字段值是否小于 99 且大于 10,并仅输出那些行.
looks a much better fit for this task. It uses a comma as a field separator, and checks if the first field value is less than 99 and bigger than 10, and outputs only those lines.
查看在线演示:
#!/bin/bash
s='0,M,19,finnish,english swedish german
9,M,30,urdu,english
122,F,26,finnish,english swedish german
83,M,20,finnish,english french swedish
44,F,20,finnish,english swedish
10,F,29,finnish,english'
awk -F, '$1 < 99 && $1 > 10' <<< "$s"
echo "---"
grep -E '^(1[1-9]|[2-8][0-9]|9[0-8]),' <<< "$s"
输出:
83,M,20,finnish,english french swedish
44,F,20,finnish,english swedish
---
83,M,20,finnish,english french swedish
44,F,20,finnish,english swedish
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