More election popular-vote numbers:
Statistical Modeling, Causal Inference, and Social Science: The Democrats in 2006 compared to the Republicans in 1994: The Democrats' victory in the 2006 election has been compared to the Republicans' in 2004. But the Democrats actually did a lot better in terms of the vote. The Democrats received 56% of the average district vote for the two parties in 2006, whereas the Republicans only averaged 51.6% in 1994.... The 2006 outcome of 56% for the Democrats is comparable to their typical vote shares as the majority party in the decades preceding the 1994 realignment....
Even with their large vote majority, the Democrats only received 53.3% of the seats in the House. This is as we and Bafumi et al. anticipated. More info on the seats-votes relationship is in our recent paper. (For example, had the Republicans received 56% of the vote in 2006, we estimate they would've won about 250 seats.)
By the way, the Democrats' 56% share of the two-party vote tracks closely with the "generic congressional vote" in which they were also getting 56% in the polls (that is, 52.1%/(52.1% + 40.6%)).
I ddidn't have a spreadsheet with all the 2006 data, but the New York Times reported that the average swing was 7.6%. We actually calculate average district votes by imputing 75% for uncontested races (to represent the strength that the party might have had if the district had been contested; the 75% comes from computing the average vote in districts just before and just after being uncontested, based on historical data), so we needed to make corrections for uncontesteds.
In 2004, we have 31 uncontested Democrats and 37 uncontested Republicans: the average district vote for the Democrats was 50.5% using our correction (or 50.0% if you simply plug in 100% for uncontested races). In 2006, the NYT showed 30 uncontested Democrats and 4 uncontested Republicans, yielding an average district vote of 56.1% using our correction (or 57.6% if you simply plug in 100%). The 57.6% number is really dramatic but I think it overstates the Democrats' strength in giving them 100% in all those districts. [John K. has a slightly different count of the number of uncontesteds, but it wouldn't change the result much.]
Another option is to use total vote rather than average district vote. We discuss this in Section 3.3 of our paper (in particular, see Figure 4). The short answer is that we use average district vote because it represents total support for the parties across the country. The Democrats tend to do better in lower-turnout districts and so their total vote is typically slightly lower than their average district vote.