Rails:循环比较两个数组并将匹配推送到新数组
我有两个数组，我需要互相检查。当找到匹配时，我需要从第一个数组中取消值，并将该值推送到一个新数组。
下面是一个示例：
第一阵列import_data
["mario", "blue", "SGC", "CJ672PA0"],
["solid snake", "green", "NID", "VI6965KD"],
["samus", "maroon", "TRUST", "DNYU6539"],
["deckard cain", "purple", "JAFA", "SJW252MZ"],
["wedge", "yellow", "WALTER", "UJ28NVM1"]第二阵列current_data
["gordon", "orange", "DRONE", "OGJR8C0D"],
["samus", "black", "TRUST", "DNYU6539"],
["commander shepard", "red", "WHX", "TH985OI3"]请注意，"samus“显示在两个数组中，所有数据都相同，但颜色发生了更改。
我的意图是循环遍历import_data，检查数组中每个项的最后一个索引(8位数字的"id")，并将其与current_data进行比较。每次找到“匹配”(基于8位数字的"id")时，我都希望将该项推入一个新的数组update_data中，然后在import_data数组中进行匹配的unset。
问题是两个数组的长度不同。因此，一旦我的循环超过一项的长度，它就不能再检查该索引，从而导致错误。
我的代码本质上是这样的：
import_data.each_with_index do |data, index|
if data[3] == current_data[index].id
# @update_data << somehow push all of the data
end
end之所以不能工作，是因为import_data的长度比current_data的长。这将导致错误：
“未定义的方法‘`id’for nil:NilClass"
理想情况下，最终结果如下：
"samus“已从import_data数组中移除并推送到update_data数组中。
import_data阵列一个又一个循环
["mario", "blue", "SGC", "CJ672PA0"],
["solid snake", "green", "NID", "VI6965KD"],
---
["deckard cain", "purple", "JAFA", "SJW252MZ"],
["wedge", "yellow", "WALTER", "UJ28NVM1"]新填充数组update_data
["samus", "maroon", "TRUST", "DNYU6539"]回答 2
Stack Overflow用户
发布于 2015-03-24 02:32:00
由于普拉卡什的答案是可行的，我认为这是比较直接的：
import_data = [
["mario", "blue", "SGC", "CJ672PA0"],
["solid snake", "green", "NID", "VI6965KD"],
["samus", "maroon", "TRUST", "DNYU6539"],
["deckard cain", "purple", "JAFA", "SJW252MZ"],
["wedge", "yellow", "WALTER", "UJ28NVM1"]
]
current_data = [
["gordon", "orange", "DRONE", "OGJR8C0D"],
["samus", "black", "TRUST", "DNYU6539"],
["commander shepard", "red", "WHX", "TH985OI3"]
]
current_ids = current_data.map {|c| c[3]}
update_data = import_data.inject([]) do |result, data|
result << import_data.delete(data) if current_ids.include?(data[3])
result
end跑完之后你会得到：
update_data = [
["samus", "maroon", "TRUST", "DNYU6539"]
]
import_data = [
["mario", "blue", "SGC", "CJ672PA0"],
["solid snake", "green", "NID", "VI6965KD"],
["deckard cain", "purple", "JAFA", "SJW252MZ"],
["wedge", "yellow", "WALTER", "UJ28NVM1"]
]希望能帮上忙。
Stack Overflow用户
发布于 2015-03-23 23:38:08
完成你正在做的事情的一种方法：
import_data = [
["mario", "blue", "SGC", "CJ672PA0"],
["solid snake", "green", "NID", "VI6965KD"],
["samus", "maroon", "TRUST", "DNYU6539"],
["deckard cain", "purple", "JAFA", "SJW252MZ"],
["wedge", "yellow", "WALTER", "UJ28NVM1"]
]
current_data = [
["gordon", "orange", "DRONE", "OGJR8C0D"],
["samus", "black", "TRUST", "DNYU6539"],
["commander shepard", "red", "WHX", "TH985OI3"]
]查找这两个数组的键：
import_data_keys = import_data.map { |x| x[3] }
=> ["CJ672PA0", "VI6965KD", "DNYU6539", "SJW252MZ", "UJ28NVM1"]
current_data_keys = current_data.map { |x| x[3] }
=> ["OGJR8C0D", "DNYU6539", "TH985OI3"]通过将这两个数组相交来查找公共密钥：
common_keys = import_data_keys & current_data_keys
=> ["DNYU6539"]使用common_keys从import_data中删除并推送到update_data。
common_keys.each do |key|
import_data.each_with_index do |data, index|
if data[3] == key
update_data << import_data[index]
import_data.delete_at(index)
break
end
end
end注意：break语句用于在元素找到和处理后停止该迭代。
https://stackoverflow.com/questions/29221948
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