Question

我从xlsx文件中提取数据，如下所示：

[6, 2]=>"Comment", [6, 3]=>"Complaint", [6, 4]=>"Enquiry"

我想搜索数组并返回括号中第二个数字的值。即，搜索Complaint应该返回3。任何帮助表示赞赏。

NB - 根据对数组的评论进行了更新。在OP中删除了字符串转换。

require 'roo'

today_file=(File.dirname(__FILE__) + '/output/today-report.xlsx')

def by_value(data, value)
  found = data.find do |k, v|
    v == value
  end

  found and found[0][1]
end

data = Roo::Excelx.new (today_file)

output = by_value(data, "Complaint").inspect

puts output

运行时，返回＆＃39; nil＆＃39;

此数组的输出采用以下形式：

{[1, 1]=>"YESTERDAY - 02/10/14", [1, 4]=>"Another comment below the fold - scroll down", [2, 1]=>"Yesterday we received 11 regarding the service.", [4, 1]=>"TYPE", [6, 1]=>"", [6, 2]=>"Comment", [6, 3]=>"Complaint", [6, 4]=>"Enquiry", [6, 5]=>"Total (Type)", [7, 1]=>"Editorial and other queries", [7, 2]=>1.0, [7, 3]=>7.0,...}

Answer 1

如果您是Ruby新手，那么您需要花时间熟悉Enumerable库。它有大量用于操作和提取数据的工具。

在这种情况下，find完成工作：

def by_value(data, value)
  # Find the key/value pair where the value matches
  found = data.find do |k, v|
    v == value
  end

  # If one of these was found, pull the second number from the key
  found and found[0][1]
end

data = {[6, 2]=>"Comment", [6, 3]=>"Complaint", [6, 4]=>"Enquiry"}

puts by_value(data, "Complaint").inspect
# => 3

值得注意的是，由于您正在进行线性搜索，因此这是一个相对较慢的操作。如果这种情况经常发生，你可以将其反转，以便更快地进行搜索：

def by_value(data, value)
  found = data[value]

  found and found[1]
end

# Using Hash#invert switches the keys and values so you can now look-up by value
data = {[6, 2]=>"Comment", [6, 3]=>"Complaint", [6, 4]=>"Enquiry"}.invert

puts by_value(data, "Complaint").inspect
# => 3

Answer 2

这是一个奇怪的要求。这是怎么做的。

word = "Complaint"
data_string[/, (\d+)\]=>"#{Regexp.escape(word)}"/, 1] # => "3"

在字符串中的单词之前查找字符

2 个答案: