United States presidential election in Rhode Island, 1948

Main article: United States presidential election, 1944

United States presidential election in Rhode Island, 1948

November 2, 1948

x195px Nominee Harry S. Truman Thomas E. Dewey Party Democratic Republican Home state Missouri New York Running mate Alben William Barkley Earl Warren Electoral vote 4 0 Popular vote 188,736 135,787 Percentage 57.59% 41.44%

President before election Harry S. Truman Democratic Elected President Harry S. Truman Democratic

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.

Rhode Island was won by Democratic candidate, incumbent President Harry S. Truman over Republican candidate New York governor Thomas E. Dewey.

Truman won Rhode Island by a margin of 16.15 percent.

Results

United States presidential election in Rhode Island, 1948
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 Ohio	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%