I have a Dictionary of dictionaries :
SortedDictionary<int,SortedDictionary<string,List<string>>>
I would like to merge two dictionaries for example key = 2 and key = 3.
Very important I could have duplicate keys in the respective dictionaries.
example key = 2 has a dictionary Key,Value "1000",{a,b,c}
"1000",{z}
.
So I would like to merge key 2 and key 3 and the result would be the following Sorted Dictionary key, Value 开发者_C百科"1000",{a,b,c,z}
.
I am a beginner with LINQ syntax so could you help solve this with detail code..
Thanks
This is not a LINQ problem. Here's a generic version that will solve your problem.
static class SortedDictionaryExtensions {
public static void MergeKeys<TKey1, TKey2, TValue>(
this SortedDictionary<TKey1, SortedDictionary<TKey2, List<TValue>>> dictionary,
TKey1 intoKey,
TKey1 fromKey
) {
if (dictionary == null) {
throw new ArgumentNullException("dictionary");
}
if (intoKey == null) {
throw new ArgumentNullException("intoKey");
}
if (fromKey == null) {
throw new ArgumentNullException("fromKey");
}
SortedDictionary<TKey2, List<TValue>> to;
SortedDictionary<TKey2, List<TValue>> from;
if (!dictionary.TryGetValue(intoKey, out to)) {
throw new ArgumentOutOfRangeException("intoKey");
}
if (!dictionary.TryGetValue(fromKey, out from)) {
throw new ArgumentOutOfRangeException("fromKey");
}
foreach(TKey2 key in from.Keys) {
if (to.Keys.Contains(key)) {
to[key].AddRange(from[key]);
}
else {
to.Add(key, from[key]);
}
}
dictionary.Remove(fromKey);
}
}
Usage:
SortedDictionary<int, SortedDictionary<string, List<string>>> list =
new SortedDictionary<int, SortedDictionary<string, List<string>>>();
list.Add(2, new SortedDictionary<string, List<string>>());
list[2].Add("1000", new List<string>() { "a", "b", "c" });
list[2].Add("2000", new List<string>() { "b", "c" });
list.Add(4, new SortedDictionary<string, List<string>>());
list[4].Add("1000", new List<string>() { "z" });
list[4].Add("3000", new List<string>() { "y" });
list.MergeKeys(2, 4);
Here's how you approach a problem like this. First, specify what you're trying to do.
Given a SortedDictionary<TKey1, SortedDictionary<TKey2, List<TValue>>>
and two keys intoKey
and fromKey
in the dictionary, merge the dictionary with key fromKey
into the dictionary with key intoKey
.
Now specify what it means to merge two dictionaries. Given two dictionaries to
and from
of type SortedDictionary<TKey2, List<TValue>>
to merge them means the following. For each TKey2 key
in from
there are two possiblities:
key
is into
. In this case add the listfrom[key]
to the listto[key]
.key
is not into
. In this case addkey
toto
with valuefrom[key]
.
Then, remove key fromKey
from the dictionary.
Let's translate this to code:
Given a
SortedDictionary<TKey1, SortedDictionary<TKey2, List<TValue>>>
and two keysintoKey
andfromKey
in the dictionary
SortedDictionary<TKey2, List<TValue>> to;
SortedDictionary<TKey2, List<TValue>> from;
// check that dictionary has intoKey
if (!dictionary.TryGetValue(intoKey, out to)) {
throw new ArgumentOutOfRangeException("intoKey");
}
// check that dictionary has fromKey
if (!dictionary.TryGetValue(fromKey, out from)) {
throw new ArgumentOutOfRangeException("fromKey");
}
For each
TKey2 key
infrom
there are two possiblities:
foreach(TKey2 key in from.Keys) {
// key is in to
if (to.Keys.Contains(key)) {
// add the list from[key] to the list to[key]
to[key].AddRange(from[key]);
}
// key is not in to
else {
// add an entry (key, from[key]) to the dictionary
to.Add(key, from[key]);
}
}
Then, remove key
fromKey
from the dictionary.
dictionary.Remove(fromKey);
The rest of the code is just error checking
LINQ will not help much here.
You should loop through each KeyValuePair in the second dictionary, call TryGetValue
to find the corresponding sub-dictionary for the key in the first dictionary (if it's not there, add it), then repeat the process for the sub-dictionary, and finally, add all of the items in the list from the second dictionary to the corresponding list in the first dictionary.
You should be able to convert that to C# relatively easily; we will not write all of your code for you.
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