如何使用grep从当前行中减去前一行中的数字？

Question

我使用 grep 来获取mysqladminas的输出

sudo mysqladmin ext -i10 | grep 'buffer_pool_pages_flushed'

并且输出是连续的（每 10 秒）为

| Innodb_buffer_pool_pages_flushed                             | 265708726                                        |
| Innodb_buffer_pool_pages_flushed                             | 265735665                                        |
| Innodb_buffer_pool_pages_flushed                             | 265751712                                        |
| Innodb_buffer_pool_pages_flushed                             | 265754576                                        |
| Innodb_buffer_pool_pages_flushed                             | 265774380                                        |

如何调整grep命令以输出第二列中连续数字之间的差异，例如

26939 (265735665-265708726)
16047 (265751712-265735665)
2864 (265754576-265751712)
19804 (265774380-265754576)

Answer 1

附加：

| awk '{if(NR>1){print $4-last,"("$4"-"last")"} last=$4}'

输出：

26939 (265735665-265708726)
16047 (265751712-265735665)
2864 (265754576-265751712)
19804（265774380-265754576）