1948 Iowa gubernatorial election
November 2, 1948
 
Nominee William S. Beardsley Carroll O. Switzer
Party Republican Democratic
Popular vote 553,900 434,432
Percentage 55.68% 43.67%
County results
Beardsley:      50–60%      60–70%      70–80%
Switzer:      50–60%
Governor before election

Robert D. Blue
Republican

 Elected Governor

William S. Beardsley
Republican

The 1948 Iowa gubernatorial election was held on November 2, 1948. Republican nominee William S. Beardsley defeated Democratic nominee Carroll O. Switzer with 55.68% of the vote.

Primary elections

Primary elections were held on June 7, 1948.[1]

Democratic primary

Candidates

Results

Democratic primary results[1]
Party Candidate Votes %
Democratic Carroll O. Switzer 56,195 100.00
Total votes 56,195 100.00

Republican primary

Candidates

Results

Republican primary results[1]
Party Candidate Votes %
Republican William S. Beardsley 189,938 59.78
Republican Robert D. Blue (incumbent) 127,771 40.22
Total votes 317,709 100.00

General election

Candidates

Major party candidates

  • William S. Beardsley, Republican
  • Carroll O. Switzer, Democratic

Other candidates

  • C. E. Bierderman, Progressive
  • Marvin Galbreath, Prohibition
  • William F. Leonard, Socialist

Results

1948 Iowa gubernatorial election[2]
Party Candidate Votes % ±%
Republican William S. Beardsley 553,900 55.68%
Democratic Carroll O. Switzer 434,432 43.67%
Progressive C. E. Bierderman 3,570 0.36%
Prohibition Marvin Galbreath 2,458 0.25%
Socialist William F. Leonard 471 0.05%
Majority 119,468
Turnout 994,833
Republican hold Swing

References

  1. 1 2 3 "Summary of Official Canvass of Votes Cast in Iowa Primary Election" (PDF). Secretary of State of Iowa. 1948. Retrieved April 9, 2020.
  2. ↑ "Summary of Official Canvass of Votes Cast in Iowa General Election" (PDF). Secretary of State of Iowa. 1948. Retrieved April 9, 2020.
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