1948 Iowa gubernatorial election explained

Election Name:1948 Iowa gubernatorial election
Country:Iowa
Flag Image:Flag of Iowa (variant).svg
Type:Presidential
Ongoing:no
Previous Election:1946 Iowa gubernatorial election
Previous Year:1946
Next Election:1950 Iowa gubernatorial election
Next Year:1950
Election Date:November 2, 1948
Nominee1:William S. Beardsley
Party1:Republican Party (United States)
Popular Vote1:553,900
Percentage1:55.68%
Nominee2:Carroll O. Switzer
Party2:Democratic Party (United States)
Popular Vote2:434,432
Percentage2:43.67%
Map Size:240px
Governor
Before Election:Robert D. Blue
Before Party:Republican Party (United States)
After Election:William S. Beardsley
After Party:Republican Party (United States)

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

Republican primary

Candidates

Results

General election

Candidates

Major party candidates

Other candidates

Results

Notes and References

  1. Web site: Summary of Official Canvass of Votes Cast in Iowa Primary Election . . 1948 . April 9, 2020.