如何使用egrep计算文件中整数的数量？
我试图将其作为模式发现问题来解决。实际上，我面临着如何表示字符范围 [0-9] 的问题不断地其中包括开头之前的“空格”和结尾之后的“空格或点”。我认为后者可以分别使用 \ 来解决。另外，它之间不应包含点，否则它不会是整数。我无法使用可用的工具和技术将此逻辑转换为正则表达式。
My name is 2322.
33 is my sister.
I am blessed with a son named 55.
Why are you so 69. Is everything 33.
66.88 is not an integer
55whereareyou?
正确答案应该是 5，即 2322、33、55、69 和 33。
grep -Eo '(^| )([0-9]+[\.\?\=\:]?( |$))+' | wc -w
^^ ^ ^ ^ ^ ^ ^
|| | | | | | |
E = extended regex--------+| | | | | | |
o = extract what found-----+ | | | | | |
starts with new line or space---+ | | | | |
digits--------------------------------+ | | | |
optional dot, question mark, etc.-------------+ | | |
ends with end line or space----------------------------+ | |
repeat 1 time or more (to detect integers like "123 456")--+ |
count words------------------------------------------------------+
注：123. 123？ 123：也算作整数
Test:
#!/bin/bash
exec 3<<EOF
My name is 2322.
33 is my sister.
I am blessed with a son named 55.
Why are you so 69. Is everything 33.
66.88 is not an integer
55whereareyou?
two integers 123 456.
how many tables in room 400? 50.
50? oh I thought it was 40.
23: It's late, 23:00 already
EOF
grep -Eo '(^| )([0-9]+[\.\?\=\:]?( |$))+' <&3 | \
tee >(sleep 0.5; echo -n "integer counted: "; wc -w; )
Outputs:
2322.
33
55.
69.
33.
123 456.
400? 50.
50?
40.
23:
integer counted: 12
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