Jay Bruce and the Mets have reunited, and it feels so good.
Bruce and the Mets agreed to terms on a $39 million contract last week, according to a report by MLB.com's Mark Feinsand. MLB Network insider Jon Heyman reported that the deal includes a partial five-team no-trade clause. The club officially announced the deal Tuesday night but has not confirmed the terms of the contract.
"The culture here is something that is very important to me," said Bruce at a news conference Wednesday. "The Mets came in and were very direct. ... They wasted no time in getting down to business and showing that they wanted me back. They wanted me to be here and wanted me to be a part of what I consider to be some unfinished business here as a Met."
"I can't tell you how happy we are to have him back," said Mets general manager Sandy Alderson. "Jay came to us in 2016, played a significant role in the late-season qualification for the playoffs and had an exceptional year for us last year and went on to perform extremely well for Cleveland. Strictly on the basis of performance, we're very glad to have him back."
Bruce has primarily played right field in his professional career, but the Mets will ask him to play some first base this season with Dominic Smith establishing himself as an everyday big leaguer. Bruce played 11 games at first last year, all for New York.
The signing reunites Bruce with the Mets, whom he played for in parts of 2016 and '17. The three-time All-Star spent the first nine seasons of his Major League career with the Reds before he was traded to the Mets on Aug. 1, 2016. He was again dealt on Aug. 9, 2017, to the Indians in exchange for Minor League pitcher Ryder Ryan.

With the Mets and Indians in 2017, Bruce batted .254/.324/.508 with a career-high 36 home runs and 101 RBIs in 146 games. He also played in the American League Division Series for the Indians against the Yankees in October, hitting .278 with a pair of home runs and four RBIs.