文本分组文本

Question

我需要帮助在分组文本..I有一个像这样的商家列表，我们可以看到，前几个属于CENTURYLINK旁边的智能ATT ..is有一种方法，以分组/标签这些文本的单一标签或分类这些文本根据他们落入的池。

提前感谢

001 CENTURYLINK IREP

003 CENTURYLINK我的帐户

003-ClearTalk无线

004 CENTURYLINK IVR

005 CENTURYLINK循环

006 CENTURYLINK WIFI

007 CENTURYLINK电缆

111智能ATT

112智能ATT

113 -智能ATT

114智能ATT

120 -智能ATT

131 - SMART - ATT

137 - SMART - ATT

无线AMERY

无线安娜

无线APTOS

无线阿卡迪亚

无线阿诺兹标准杆

无线阿什兰

无线的雅典

Answer 1

你有几个选择。最简单的方法之一是匹配厂商的子字符串，如下所示：

import java.util.Arrays;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.TreeMap;
import java.util.stream.Collectors;

public class GroupVendors {
    public static void main(final String[] args) {
        final List<String> vendors = Arrays.asList(
            "CENTURYLINK",
            "SMART",
            "ATT",
            "A WIRELESS");

        final List<String> uncategorizedVendors = Arrays.asList(
            "001 CENTURYLINK IREP",
            "003 CENTURYLINK MY ACCOUNT",
            "003-ClearTalk Wireless",
            "004 CENTURYLINK IVR",
            "005 CENTURYLINK RECURRING",
            "006 CENTURYLINK WIFI",
            "007 CENTURYLINK CABLE",
            "111 SMART ATT",
            "112 SMART ATT",
            "113 - SMART - ATT",
            "114 SMART ATT",
            "120 - SMART - ATT",
            "131 - SMART - ATT",
            "137 - SMART - ATT",
            "A WIRELESS AMERY",
            "A WIRELESS ANNA",
            "A WIRELESS APTOS",
            "A WIRELESS ARCADIA",
            "A WIRELESS ARNOLDS PAR",
            "A WIRELESS ASHLAND",
            "A WIRELESS ATHENS");

        final Map<String, List<String>> categorizedVendors = new TreeMap<>();

        for (final String vendor : vendors) {
            categorizedVendors.put(vendor, new LinkedList<String>());
        }

        for (final String vendor : uncategorizedVendors) {
            for (final Map.Entry<String, List<String>> entry : categorizedVendors.entrySet()) {
                final String category = entry.getKey();
                if (vendor.contains(category)) {
                    final List<String> bin = entry.getValue();
                    bin.add(vendor);
                }
            }
        }

        for (final Map.Entry<String, List<String>> entry : categorizedVendors.entrySet()) {
            final String category = entry.getKey();
            final List<String> bin = entry.getValue();
            System.out.printf("vendors(\"%s\") = {%n", category);
            if (!bin.isEmpty()) {
                System.out.printf("    %s%n",
                    bin.stream()
                        .map((vendor) -> String.format("\"%s\"", vendor))
                        .collect(Collectors.joining(",\n    ")));
            }
            System.out.println("}");
        }
    }
}

示例运行：

% java GroupVendors
vendors("A WIRELESS") = {
    "A WIRELESS AMERY",
    "A WIRELESS ANNA",
    "A WIRELESS APTOS",
    "A WIRELESS ARCADIA",
    "A WIRELESS ARNOLDS PAR",
    "A WIRELESS ASHLAND",
    "A WIRELESS ATHENS"
}
vendors("ATT") = {
    "111 SMART ATT",
    "112 SMART ATT",
    "113 - SMART - ATT",
    "114 SMART ATT",
    "120 - SMART - ATT",
    "131 - SMART - ATT",
    "137 - SMART - ATT"
}
vendors("CENTURYLINK") = {
    "001 CENTURYLINK IREP",
    "003 CENTURYLINK MY ACCOUNT",
    "004 CENTURYLINK IVR",
    "005 CENTURYLINK RECURRING",
    "006 CENTURYLINK WIFI",
    "007 CENTURYLINK CABLE"
}
vendors("SMART") = {
    "111 SMART ATT",
    "112 SMART ATT",
    "113 - SMART - ATT",
    "114 SMART ATT",
    "120 - SMART - ATT",
    "131 - SMART - ATT",
    "137 - SMART - ATT"
}

我假设您感兴趣的供应商类别列表是"CENTURYLINK“、"SMART”、"ATT“和"A WIRELESS”。这样做的效果是对在两个回收站中同时包含"SMART“和"ATT”的所有条目进行分类。如果您希望将每个供应商准确地归入一个库中，那么当类别是多余的时，您将需要确定您更喜欢哪个供应商。