The 1948 United States presidential election in Rhode Island took place on November 2, 1948, as part of the 1948 United States presidential election. State voters chose four electors to the Electoral College, which selected the president and vice president.
Rhode Island was won by Democratic candidate, incumbentPresidentHarry S. Truman of Missouri, over the Republican candidate, GovernorThomas E. Dewey of New York. Truman ran with SenatorAlben W. Barkley of Kentucky as his running mate, while Dewey ran with GovernorEarl Warren of California as his running mate.
Truman won Rhode Island by a margin of 16.15%.
Results[]
1948 United States presidential election in Rhode Island[1]
Party
Candidate
Running mate
Popular vote
Electoral vote
Count
%
Count
%
Democratic
Harry S. Truman of Missouri
Alben William Barkley of Kentucky
188,736
57.59%
4
100.00%
Republican
Thomas Edmund Dewey of New York
Earl Warren of California
135,787
41.44%
0
0.00%
Progressive
Henry Agard Wallace of Iowa
Glen Hearst Taylor of Idaho
2,619
0.80%
0
0.00%
Socialist
Norman Thomas of New York
Tucker Powell Smith of Michigan
429
0.13%
0
0.00%
Socialist Labor
Edward A. Teichert of Pennsylvania
Stephen Emery of New York
131
0.04%
0
0.00%
Total
327,702
100.00%
4
100.00%
See also[]
United States presidential elections in Rhode Island