Question

我无法弄清楚如何跳过这些数字（例如1.，2.，3。）。我不应该修改字符串，但仍需要将每行的所有首字母大写（忽略数字）。也许我可以使用each_line和sub，也许match方法。

1.     i was just doing this problem.
2. also eating so much food.
3.     it was nice listening to the Mahler.

要

1.     I was just doing this problem.
2. Also eating so much food.
3.     It was nice listening to the Mahler.

Answer 1

您可以使用gsub更新多行：

str = <<TEXT
1.     i was just doing this problem.
2. also eating so much food.
3.     it was nice listening to the Mahler.
TEXT

puts str.gsub(/^(\d+\.[ \t]+)(\w)/) { "#{$1}#{$2.upcase}" }

输出：

1.     I was just doing this problem.
2. Also eating so much food.
3.     It was nice listening to the Mahler.

^匹配行的开头
(\d+\.[ \t]+)捕获编号行的开头，即一个或多个数字，后跟一个文字点和空格/制表符
(\w)捕获单个字符

第一个捕获组未经修改返回：#{$1}，而第二个捕获组是upcased：#{$2.upcase}

由于upcase仅影响字母，您还可以upcase包括第一个字母在内的所有内容：

puts str.gsub(/^\d+\.[ \t]+\w/, &:upcase)

Answer 2

使用匹配前和匹配后变量

可能有更有效的方法来执行此操作，但一种简单的方法是使用pre-match和post-match特殊变量来打印每行的未修改部分，同时仍然调用String#upcase在它的第一封信上。例如：

text = <<'EOF'
1.     i was just doing this problem.
2. also eating so much food.
3.     it was nice listening to the Mahler.
EOF

text.each_line { |line| line =~ /\p{Alpha}/; print $`, $&.upcase, $' }

这将正确打印：

1.     I was just doing this problem.
2. Also eating so much food.
3.     It was nice listening to the Mahler.

，同时仍然返回原始的未经修改的字符串：

＃=＆GT; “1.我只是在做这个问题。\ n2。还吃了那么多食物。\ n3。听马勒很好听。\ n”

Answer 3

可以预测，修改后的字符串是保留的，所以我建议像：

str=
%{1.     i was just doing this problem.
2. also eating so much food.
3.
4.     It was nice listening to the Mahler.}

new_str = str.each_line.map { |l| l.sub(/[a-z]/i) { |c| c.upcase } }.join
puts new_str
  # 1.     I was just doing this problem.
  # 2. Also eating so much food.
  # 3.
  # 4.     It was nice listening to the Mahler.

这不会改变str：

puts str
  # 1.     i was just doing this problem.
  # 2. also eating so much food.
  # 3.
  # 4.     It was nice listening to the Mahler.

或者，您可以添加捕获组并在块中引用其值：

str.each_line.map { |l| l.sub(/([a-z])/i) {  $1.upcase } }.join

大写每个编号行的第一个字母

3 个答案:

使用匹配前和匹配后变量