I have the following algorithm which retrieves information in a List of KeyValuePair (prices) and write them in a DataRow. I was wondering if I could achieve the same result without j.

if (prices != null)
   for (int i = 8, j=0; j < prices.Count; i+=2, j++) 
      res[i] = prices[j].Value;
      res[i+1] = prices[j].Key;

The ouput is a DataRow as follow :

res[8] = prices[0].Value
res[9] = prices[0].Key
res[10] = prices[1].Value
res[11] = prices[1].Key
res[12] = prices[2].Value
res[13] = prices[2].Key



2 Answers 2


You can calculate i from j as 2*j + 8 (and 2*j + 9). That would change your code to:

for (int j = 0; j < prices.Count; j++) 
   res[2*j + 8] = prices[j].Value;
   res[2*j + 9] = prices[j].Key;

Though I'm not sure whether this is actually better code than the original.

  • \$\begingroup\$ Well at least this answer my question perfectly. I dont know how I missed that however. \$\endgroup\$
    – WizLiz
    Nov 22, 2013 at 19:04

I would approach this first by flattening the prices dictionary into a sequence of values eg { Key, Value, Key, Value, Key, Value }.

// This method is returning object for now since I'm not clear what types are in your data row..
IEnumerable<object> Flatten<K, V>(IEnumerable<KeyValuePair<K, V>> pairs)
    foreach (var pair in pairs)
        yield return pair.Value;
        yield return pair.Key;

Using this I can simply loop over the flattened dictionary and copy the values into the data row:

int offset = 8;
foreach (var x in Flatten(prices))
    res[offset++] = x;

I suspect there is a far better term than "flatten" for what I'm doing with the dictionary here.

  • \$\begingroup\$ that looks a lot cleaner than what I was going to do. \$\endgroup\$
    – Malachi
    Nov 22, 2013 at 16:08
  • 2
    \$\begingroup\$ The question says it's a list of key-value pairs, not a dictionary. (Though that will just mean that you would change the type of pairs.) \$\endgroup\$
    – svick
    Nov 22, 2013 at 16:10
  • \$\begingroup\$ @svick good spot, changed to IEnumerable<KeyValuePair<K, V>>. \$\endgroup\$
    – MattDavey
    Nov 22, 2013 at 16:18
  • \$\begingroup\$ Though your answer is good I was looking for something like what svick did. Thanks anyway ! \$\endgroup\$
    – WizLiz
    Nov 22, 2013 at 19:04
  • 1
    \$\begingroup\$ That is probably faster than the accepted answer because there is no arithmetic overhead. (Note: you need to yield Value before Key to conform OP's situation) \$\endgroup\$ Nov 22, 2013 at 22:10

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