1948 Iowa gubernatorial election

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.

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%

Governor before election

Robert D. Blue
Republican

Elected Governor

William S. Beardsley
Republican

Primary elections

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

Candidates

Results

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

Candidates

Results

Republican primary results[1]
Party Candidate Votes %
Republican William S. Beardsley 189,938 59.78
Republican Robert D. Blue 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

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