1948 United States presidential election in Iowa
| ||||||||||||||||||||||||||
All 10 Iowa votes to the Electoral College | ||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| ||||||||||||||||||||||||||
County Results
| ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
.svg){kind=link}
.jpg){kind=link}
.jpg){kind=link}
| Elections in Iowa |
|---|
The 1948 United States presidential election in Iowa took place on November 2, 1948, as part of the 1948 United States presidential election. Iowa voters chose ten representatives, or electors, to the Electoral College, who voted for president and vice president.
Iowa was won by incumbent Democratic President Harry S. Truman of neighbouring Missouri, running with Senator Alben W. Barkley, with 50.31% of the popular vote, against Governor Thomas E. Dewey (R–New York), running with Governor Earl Warren, with 47.58% of the popular vote.
Defying the expectations of a Dewey sweep in a state he won four years earlier, in an unexpected twist, Truman would manage to flip the state from Republican to Democratic, a swing of roughly 6%.