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1948 United States presidential election in Rhode Island

From Wikipedia, the free encyclopedia

1948 United States presidential election in Rhode Island

← 1944 November 2, 1948 1952 →
Harry S. Truman.jpg
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

Elected President

Harry S. Truman

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.


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%


  1. ^ "1948 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.

This page was last edited on 27 July 2019, at 16:49
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