paul gordon,jin kazama,1277,1268,21-12,21-19
yoshimistu,the rock,2020,2092,21-9,21-23,25-27
... lot more
我的意思是,如何将每一行和逗号sparator删除为像这样的哈希
result = {
line_num: { name1: "paula wood", name2: "sarah carnley", m1: 1277, m2: 1268, sc1: 21, sc2: 12, sc3: 21, sc4: 19 }
}
我尝试像这样编码 即时通讯使用text2re进行正则表达式here
doc = File.read("doc.txt")
lines = doc.split("\n")
counts = 0
example = {}
player1 = '((?:[a-z][a-z]+))(.)((?:[a-z][a-z]+))'
player2 = '((?:[a-z][a-z]+))(.)((?:[a-z][a-z]+))'
re = (player1 + player2 )
m = Regexp.new(re, Regexp::IGNORECASE)
lines.each do |line|
re1='((?:[a-z][a-z]+))' # Word 1
re2='(.)' # Any Single Character 1
re3='((?:[a-z][a-z]+))' # Word 2
re4='(.)' # Any Single Character 2
re5='((?:[a-z][a-z]+))' # Word 3
re6='(.)' # Any Single Character 3
re7='((?:[a-z][a-z]+))' # Word 4
re=(re1+re2+re3+re4+re5+re6+re7)
m=Regexp.new(re,Regexp::IGNORECASE);
if m.match(line)
word1=m.match(line)[1];
c1=m.match(line)[2];
word2=m.match(line)[3];
c2=m.match(line)[4];
word3=m.match(line)[5];
c3=m.match(line)[6];
word4=m.match(line)[7];
counts += 1
example[counts] = word1+word2
puts example
end
end
# (/[a-z].?/)
但输出与我的预期不符
1=>"", 2=>"indahdelika", 3=>"masam",
..more
答案 0 :(得分:1)
您的数据以逗号分隔,因此请使用CSV类,而不是尝试滚动自己的解析器。如果您尝试使用逗号分割,有龙会等着您。
我使用:
require 'csv'
data = "paul gordon,jin kazama,1277,1268,21-12,21-19
yoshimistu,the rock,2020,2092,21-9,21-23,25-27
"
hash = {}
CSV.parse(data).each_with_index do |row, i|
name1, name2, m1, m2, sc1_2, sc3_4 = row
sc1, sc2 = sc1_2.split('-')
sc3, sc4 = sc3_4.split('-')
hash[i] = {
name1: name1,
name2: name2,
m1: m1,
m2: m2,
sc1: sc1,
sc2: sc2,
sc3: sc3,
sc4: sc4,
}
end
结果是:
hash
# => {0=>
# {:name1=>"paul gordon",
# :name2=>"jin kazama",
# :m1=>"1277",
# :m2=>"1268",
# :sc1=>"21",
# :sc2=>"12",
# :sc3=>"21",
# :sc4=>"19"},
# 1=>
# {:name1=>"yoshimistu",
# :name2=>"the rock",
# :m1=>"2020",
# :m2=>"2092",
# :sc1=>"21",
# :sc2=>"9",
# :sc3=>"21",
# :sc4=>"23"}}
由于您正在从文件中读取内容,请使用" Reading from a file a line at a time"修改上述内容。文档中的示例。
如果数字必须是整数,请将哈希定义调整为:
hash[i] = {
name1: name1,
name2: name2,
m1: m1.to_i,
m2: m2.to_i,
sc1: sc1.to_i,
sc2: sc2.to_i,
sc3: sc3.to_i,
sc4: sc4.to_i,
}
结果是:
# => {0=>
# {:name1=>"paul gordon",
# :name2=>"jin kazama",
# :m1=>1277,
# :m2=>1268,
# :sc1=>21,
# :sc2=>12,
# :sc3=>21,
# :sc4=>19},
# 1=>
# {:name1=>"yoshimistu",
# :name2=>"the rock",
# :m1=>2020,
# :m2=>2092,
# :sc1=>21,
# :sc2=>9,
# :sc3=>21,
# :sc4=>23}}
# :sc4=>"23"}}
答案 1 :(得分:0)
这是你可以做到的另一种方式。我没有假设每行的项目数量是:namex
,:scx
或:mx
的值,或者这些项的顺序。
<强>代码强>
def hashify(str)
str.lines.each_with_index.with_object({}) { |(s,i),h| h[i] = inner_hash(s) }
end
def inner_hash(s)
n = m = sc = 0
s.split(',').each_with_object({}) do |f,g|
case f
when /[a-zA-Z].*/
g["name#{n += 1}".to_sym] = f
when /\-/
g["sc#{sc += 1}".to_sym], g["sc#{sc += 1}".to_sym] = f.split('-').map(&:to_i)
else
g["m#{m += 1}".to_sym] = f.to_i
end
end
end
示例强>
str = "paul gordon,jin kazama,1277,1268,21-12,21-19
yoshimistu,the rock,2020,2092,21-9,21-23,25-27"
hashify(str)
#=> {0=>{:name1=>"paul gordon", :name2=>"jin kazama",
# :m1=>1277, :m2=>1268,
# :sc1=>21, :sc2=>12, :sc3=>21, :sc4=>19},
# 1=>{:name1=>"yoshimistu", :name2=>"the rock",
# :m1=>2020, :m2=>2092,
# :sc1=>21, :sc2=>9, :sc3=>21, :sc4=>23, :sc5=>25, :sc6=>27}
# }