List 如何在Groovy中根据列的值过滤列表中的元素

我使用的是

Groovy

，有一个

列表

，其中包含一定数量的元素：

def rList = [["Using Patient Feedback","Completed","No Results Available","Condition1","Other","Phase I","18 Years and older   (Adult, Senior)","Other","https://folder1//folder2"],["The Effect of a Education ","Completed","No Results Available","Condition1","Behavioral","Glycemic Control","18 Years and older   (Adult, Senior)","Other","https://folder1//folder3"],["The Public Private Partnership","Completed","No Results Available","Condition1","Enhanced Education","Improvement in A1C","18 Years to 85 Years   (Adult, Senior)","NIH","https://folder1//folder4"],["Evaluation of a Pilot","Completed","No Results Available","Condition1","Behavioral","Change in percent of total calories","10 Years to 19 Years   (Child, Adult)","Other","https://folder1//folder5"],["Self Management Skills","Completed","No Results Available","Condition1","Behavioral","Metabolic control","18 Years and older   (Adult, Senior)","Industry","https://folder1//folder6"]]

每个

列表

以及上面的

列表

都有

9个
元素/列。我还有一个搜索字符串，定义如下：

def fundingType = "NIH"

现在我要做的是在每个列表的
8th
元素/列中查找该字符串的存在，如果它存在，则将其余8个元素存储在8个不同的变量中-
var1
，
var2
，…
var8
。如何在
Groovy
中执行此操作？
您应该能够使用
find
轻松获得它，如下所示：

def fundingType = "NIH"
println rList.find{ fundingType == it[7] }

你可以在网上快速试用​

代替8个变量，您可以将其更改为：

def result = rList.find{ fundingType == it[7] }

使用
result[0]
，
result[1]
…而不使用额外的8个变量。
您应该能够使用
find
轻松获得它，如下所示：

def fundingType = "NIH"
println rList.find{ fundingType == it[7] }

你可以在网上快速试用​

代替8个变量，您可以将其更改为：

def result = rList.find{ fundingType == it[7] }

并使用
结果[0]
，
结果[1]
…而不使用额外的8个变量。
类似于：

def rList = [["Using Patient Feedback","Completed","No Results Available","Condition1","Other","Phase I","18 Years and older   (Adult, Senior)","Other","https://folder1//folder2"],["The Effect of a Education ","Completed","No Results Available","Condition1","Behavioral","Glycemic Control","18 Years and older   (Adult, Senior)","Other","https://folder1//folder3"],["The Public Private Partnership","Completed","No Results Available","Condition1","Enhanced Education","Improvement in A1C","18 Years to 85 Years   (Adult, Senior)","NIH","https://folder1//folder4"],["Evaluation of a Pilot","Completed","No Results Available","Condition1","Behavioral","Change in percent of total calories","10 Years to 19 Years   (Child, Adult)","Other","https://folder1//folder5"],["Self Management Skills","Completed","No Results Available","Condition1","Behavioral","Metabolic control","18 Years and older   (Adult, Senior)","Industry","https://folder1//folder6"]]
def fundingType = "NIH"
def fundingColumnIndex = 7

def fundingRow = rList.find { it[fundingColumnIndex] == fundingType }
fundingRow.remove(fundingColumnIndex)
assert fundingRow.size() == 8
def (v1, v2, v3, v4, v5, v6, v7, v8) = fundingRow
[v1, v2, v3, v4, v5, v6, v7, v8].each { println(it) }

输出：

The Public Private Partnership
Completed
No Results Available
Condition1
Enhanced Education
Improvement in A1C
18 Years to 85 Years   (Adult, Senior)
https://folder1//folder4
公私伙伴关系
完整的
没有结果
条件1
强化教育
A1C的改进
18岁至85岁（成人、老年人）
https://folder1//folder4
比如：

def rList = [["Using Patient Feedback","Completed","No Results Available","Condition1","Other","Phase I","18 Years and older   (Adult, Senior)","Other","https://folder1//folder2"],["The Effect of a Education ","Completed","No Results Available","Condition1","Behavioral","Glycemic Control","18 Years and older   (Adult, Senior)","Other","https://folder1//folder3"],["The Public Private Partnership","Completed","No Results Available","Condition1","Enhanced Education","Improvement in A1C","18 Years to 85 Years   (Adult, Senior)","NIH","https://folder1//folder4"],["Evaluation of a Pilot","Completed","No Results Available","Condition1","Behavioral","Change in percent of total calories","10 Years to 19 Years   (Child, Adult)","Other","https://folder1//folder5"],["Self Management Skills","Completed","No Results Available","Condition1","Behavioral","Metabolic control","18 Years and older   (Adult, Senior)","Industry","https://folder1//folder6"]]
def fundingType = "NIH"
def fundingColumnIndex = 7

def fundingRow = rList.find { it[fundingColumnIndex] == fundingType }
fundingRow.remove(fundingColumnIndex)
assert fundingRow.size() == 8
def (v1, v2, v3, v4, v5, v6, v7, v8) = fundingRow
[v1, v2, v3, v4, v5, v6, v7, v8].each { println(it) }

输出：

The Public Private Partnership
Completed
No Results Available
Condition1
Enhanced Education
Improvement in A1C
18 Years to 85 Years   (Adult, Senior)
https://folder1//folder4
公私伙伴关系
完整的
没有结果
条件1
强化教育
A1C的改进
18岁至85岁（成人、老年人）
https://folder1//folder4
你能检查解决方案看它是否有用吗？你能检查解决方案看它是否有用吗？为什么要先检查
中的
？
fundingType==it[7]
不够好，效率更高吗？@cfrick，谢谢你的意见。起初我只在
中使用了
，后来意识到OP指定了位置，所以最后添加了
和
条件。更新。为什么要先检查
中的
？
fundingType==it[7]
不够好，效率更高吗？@cfrick，谢谢你的意见。起初我只在
中使用了
，后来意识到OP指定了位置，所以最后添加了
和
条件。更新。