1948 Iowa gubernatorial election

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1948 Iowa gubernatorial election
Flag of Iowa (variant).svg
  1946
November 2, 1948
1950  
  Bill Beardsley.png 3x4.svg
Nominee William S. Beardsley Carroll O. Switzer
Party Republican Democratic
Popular vote553,900434,432
Percentage55.68%43.67%

1948 Iowa gubernatorial election results map by county.svg
County results
Beardsley:     40–50%     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.

Contents

Primary elections

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

Democratic primary

Candidates

Results

Democratic primary results [1]
PartyCandidateVotes%
Democratic Carroll O. Switzer 56,195 100.00
Total votes56,195 100.00

Republican primary

Candidates

Results

Republican primary results [1]
PartyCandidateVotes%
Republican William S. Beardsley 189,938 59.78
Republican Robert D. Blue (incumbent)127,77140.22
Total votes317,709 100.00

General election

Candidates

Major party candidates

Other candidates

Results

1948 Iowa gubernatorial election [2]
PartyCandidateVotes%±%
Republican William S. Beardsley 553,900 55.68%
Democratic Carroll O. Switzer 434,43243.67%
Progressive C. E. Bierderman3,5700.36%
Prohibition Marvin Galbreath2,4580.25%
Socialist William F. Leonard4710.05%
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.