我有一个RSS源,每天输出约100篇文章.我想过滤它只包括更流行的链接,也许过滤它到50或更少.回到当天,我相信你可以使用"postrank"来做到这一点,但现在在Google收购后已经不复存在.
任何人都知道如何过滤特定的RSS源以仅包含更受欢迎的输出？
谢谢!
我有一台新机器并且安装了rbenv(我之前总是使用rvm).这个宝石现在不会加载到我的应用程序中.我没有更改代码.现在和现在的唯一区别是rbenv.
Loading development environment (Rails 5.2.1)
irb(main):001:0> SportsApi::Fetcher::Score::NBA
Traceback (most recent call last):
1: from (irb):1
NameError (uninitialized constant SportsApi::Fetcher)
有趣的是,如果我将gem克隆到我的本地机器,然后更改Gemfile中的路径,它就会加载/工作.
# gem 'sports_api', git: 'git@github.com:mikesilvis/sports_api.git'
gem 'sports_api', path: '~/dev/sports_api'
irb(main):001:0> SportsApi::Fetcher::Score::NBA
=> SportsApi::Fetcher::Score::NBA
为什么它在本地路径上工作而在git路径上不工作？关于我可以做些什么来解决这个问题(使用git路径)的任何想法？先感谢您.
编辑:这是完整的回溯.
uninitialized constant SportsApi::Fetcher
/Users/myusername/.rbenv/versions/2.5.3/lib/ruby/gems/2.5.0/gems/bootsnap-1.3.2/lib/bootsnap/load_path_cache/core_ext/active_support.rb:74:in `block in load_missing_constant'
/Users/myusername/.rbenv/versions/2.5.3/lib/ruby/gems/2.5.0/gems/bootsnap-1.3.2/lib/bootsnap/load_path_cache/core_ext/active_support.rb:8:in `without_bootsnap_cache'
/Users/myusername/.rbenv/versions/2.5.3/lib/ruby/gems/2.5.0/gems/bootsnap-1.3.2/lib/bootsnap/load_path_cache/core_ext/active_support.rb:74:in `rescue in load_missing_constant'
/Users/myusername/.rbenv/versions/2.5.3/lib/ruby/gems/2.5.0/gems/bootsnap-1.3.2/lib/bootsnap/load_path_cache/core_ext/active_support.rb:56:in `load_missing_constant'
(irb):2:in `irb_binding'
/Users/myusername/.rbenv/versions/2.5.3/lib/ruby/2.5.0/irb/workspace.rb:85:in `eval'
/Users/myusername/.rbenv/versions/2.5.3/lib/ruby/2.5.0/irb/workspace.rb:85:in `evaluate'
/Users/myusername/.rbenv/versions/2.5.3/lib/ruby/2.5.0/irb/context.rb:380:in `evaluate'
/Users/myusername/.rbenv/versions/2.5.3/lib/ruby/2.5.0/irb.rb:491:in `block (2 levels) in eval_input'
/Users/myusername/.rbenv/versions/2.5.3/lib/ruby/2.5.0/irb.rb:623:in `signal_status'
/Users/myusername/.rbenv/versions/2.5.3/lib/ruby/2.5.0/irb.rb:488:in `block in eval_input'
/Users/myusername/.rbenv/versions/2.5.3/lib/ruby/2.5.0/irb/ruby-lex.rb:246:in `block (2 levels) in each_top_level_statement'
/Users/myusername/.rbenv/versions/2.5.3/lib/ruby/2.5.0/irb/ruby-lex.rb:232:in …我有3个相等长度的数组.有些地方是零,这使事情变得复杂,但我需要保留他们的秩序.
a = [5.2, 3.0, 1.21, 7.0, 5.0, 5.0, 6.0, 8.0, 10.0, 10.0]
b = [nil, nil, [{"price"=>1.99, "size"=>269.897475661239}], nil, nil, nil, nil, nil, nil, nil]
x = [6.0, 6.2, 2.5, 5.0, 9.0, 2.36, 15.5, 20.0, nil, nil]
(第一步,我想迭代b以便b = [nil, nil, 1.99, nil, nil, nil, nil, nil, nil, nil].只需要["价格"],忽略["大小"].无法弄明白.)
第二步,我要创建一个新的数组(c),其平均a和b,但那里是零,只需要具有价值的一个.换句话说,c将= [5.2, 3.0, 1.6, 7.0, 5.0, 5.0, 6.0, 8.0, 10.0, 10.0]它看起来像a除了所述第三点的平均的1.21和1.99(1.6).
所以我有我原来的第三个阵列x …