1948 United States presidential election in Rhode Island
|Elections in Rhode Island|
The 1948 United States presidential election in Rhode Island took place on November 7, 1944, 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.
Truman won Rhode Island by a margin of 16.15 percent.
|1948 United States presidential election in Rhode Island|
|Party||Candidate||Running mate||Popular vote||Electoral vote|
|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%|
- "1948 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.
|This Rhode Island elections-related article is a stub. You can help Wikipedia by expanding it.|