Lakers forward LeBron James and Bucks guard Jrue Holiday have been named the NBA’s players of the week, the league announced (via Twitter).
James led the Lakers to a 3-1 week with averages of 35.0 PPG, 9.0 RPG, 7.0 APG and 1.3 BPG on .510/.323/.842 shooting. He continues to play at an incredibly high level at 38 years old, having won the award a couple of weeks ago as well. The Lakers are currently 22-25, the West’s No. 12 seed, but they’re only 2.5 games back of the No. 5 seed Mavericks.
Holiday had an outstanding week himself, leading Milwaukee to a 2-1 record with Giannis Antetokounmpo and Khris Middleton sidelined. He averaged 33.3 PPG, 4.7 RPG, 9.3 APG and 1.7 SPG on a stellar .569/.478/1.000 shooting slash line. The Bucks are currently 29-17, the East’s No. 3 seed, and are expected to get both of Holiday’s aforementioned teammates back on Monday.
According to the NBA (Twitter link), the other nominees in the West were Mikal Bridges, Shai Gilgeous-Alexander, Kawhi Leonard and Jamal Murray, while Joel Embiid, Darius Garland, Kyrie Irving, Dejounte Murray and Fred VanVleet were nominated in the East.
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