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1948 Iowa gubernatorial election

From Wikipedia, the free encyclopedia

1948 Iowa gubernatorial election

← 1946November 2, 19481950 →
 
NomineeWilliam S. BeardsleyCarroll O. Switzer
PartyRepublicanDemocratic
Popular vote553,900434,432
Percentage55.68%43.67%

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

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The1948 Iowa gubernatorial election was held on November 2, 1948.Republican nomineeWilliam S. Beardsley defeatedDemocratic nomineeCarroll O. Switzer with 55.68% of the vote.

Primary elections

[edit]

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

Democratic primary

[edit]

Candidates

[edit]

Results

[edit]
Democratic primary results[1]
PartyCandidateVotes%
DemocraticCarroll O. Switzer56,195100.00
Total votes56,195100.00

Republican primary

[edit]

Candidates

[edit]

Results

[edit]
Republican primary results[1]
PartyCandidateVotes%
RepublicanWilliam S. Beardsley189,93859.78
RepublicanRobert D. Blue (incumbent)127,77140.22
Total votes317,709100.00

General election

[edit]

Candidates

[edit]

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

[edit]
1948 Iowa gubernatorial election[2]
PartyCandidateVotes%±%
RepublicanWilliam S. Beardsley553,90055.68%
DemocraticCarroll O. Switzer434,43243.67%
ProgressiveC. E. Bierderman3,5700.36%
ProhibitionMarvin Galbreath2,4580.25%
SocialistWilliam F. Leonard4710.05%
Majority119,468
Turnout994,833
RepublicanholdSwing

References

[edit]
  1. ^abc"Summary of Official Canvass of Votes Cast in Iowa Primary Election"(PDF).Secretary of State of Iowa. 1948. RetrievedApril 9, 2020.
  2. ^"Summary of Official Canvass of Votes Cast in Iowa General Election"(PDF).Secretary of State of Iowa. 1948. RetrievedApril 9, 2020.
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