The Milwaukee Brewers announced they have agreed to terms with 1B/OF Darin Ruf on a contract. Ruf offers versatility and has previously played for the Philadelphia Phillies, San Francisco Giants, and New York Mets.
Ruf was acquired by the Mets last year in a trade from the Giants. His time with New York was underwhelming though, which led to the Mets ultimately deciding to move on. San Francisco re-signed Ruf this season, but he later elected MLB free agency after being waived by the team. After all of this, Ruf has landed with the Brewers.
Ruf, however, began his career with the Phillies back in 2012. He displayed promise in 2013 after hitting 14 home runs with an .806 OPS. But he failed to see consistent playing time in Philadelphia and never appeared in more than 106 games in a single season. Following the 2016 campaign, Ruf found himself out of MLB. He later returned in 2020 after inking a contract with the Giants.
Fast-forward to 2023, and Ruf had hit .261/.370/.348 with a .718 OPS across 27 plate appearances with San Francisco. Perhaps he can improve his numbers with a potential steady role in Milwaukee with the Brewers.
Milwaukee currently leads the National League Central division. They could use extra offensive help though. As of this story's publication, the Brewers rank just 16th in batting average, 17th in OBP, 18th in slugging percentage, and 14th in runs scored. It will be interesting to see how the ball club plans to utilize Darin Ruf.
This move will be overlooked by many, but could pan out for Milwaukee and Ruf.
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