我正在处理下面的问题,我需要将一系列的命令放平:
例如-下面是一个输入:
[
{'a':
{'b':
{'c':
{'d':'e'}
}
}
},
{'a':{'b':{'c':{'d':{'e':'f'}}}}},
{'a':'b'}
]
输出为:
[
{'a_b_c_d':'e'},
{'a_b_c_d_e':'f'},
{'a':'b'}
]
下面是我能想到的。有没有更好的方法来解决这个问题?
private static List<Map<String, String>> flatDictionary(final List<Map<String, Object>> input) {
List<Map<String, String>> listHolder = new ArrayList<>();
if(input == null || input.isEmpty()) {
return listHolder;
}
for(Map<String, Object> mapHolder : input) {
Map<String, String> m = new HashMap<>();
StringBuilder sb = new StringBuilder();
Map<String, String> output = helper(mapHolder, sb, m);
listHolder.add(output);
}
return listHolder;
}
private static Map<String, String> helper(final Map<String, Object> map, final StringBuilder sb, final Map<String, String> output) {
String mapValue = null;
for(Map.Entry<String, Object> holder : map.entrySet()) {
String key = holder.getKey();
Object value = holder.getValue();
if(value instanceof Map) {
sb.append(key).append("_");
helper((HashMap<String, Object>) value, sb, output);
} else if(value instanceof String) {
sb.append(key);
mapValue = (String) value;
}
output.put(sb.toString(), mapValue);
}
return output;
}
最佳答案
我会用递归。
首先定义一种方法来展平单个Map
public static void flatten(final String keyPrefix, final Map<String, Object> input, final Map<String, Object> output) {
for (final Map.Entry<String, Object> e : input.entrySet()) {
final var key = keyPrefix.isBlank() ? e.getKey() : keyPrefix + "_" + e.getKey();
if (e.getValue() instanceof Map) {
// if the nested Map is of the wrong type bad things may happen
flatten(key, (Map<String, Object>) e.getValue(), output);
} else {
output.put(key, e.getValue());
}
}
}
注意:这不会试图处理重复的密钥。
用途:
public static void main(final String[] args) throws InterruptedException {
final var data = Map.of(
"A", Map.of("a", "Expect A_a"),
"B", Map.of("b1", Map.of(
"bb1", "expect B_b1_bb1",
"bb2", "expect B_b1_bb2"
)),
"C", "Expect C");
final var output = new HashMap<String, Object>();
flatten("", data, output);
output.forEach((k, v) -> System.out.printf("%s -> %s%n", k, v));
}
输出:
C -> Expect C
A_a -> Expect A_a
B_b1_bb2 -> expect B_b1_bb2
B_b1_bb1 -> expect B_b1_bb1
现在,只需定义一个循环的方法来获取
List
public static final Map<String, Object> flattenAll(final List<Map<String, Object>> input) {
final var output = new HashMap<String, Object>();
input.forEach(map -> flatten("", map, output));
return output;
}
注意:这不会试图处理重复的密钥。
关于java - 如何在Java中拼合字典数组?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/54314503/