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 weeks ago as well. The Lakers are currently 22-25, the West’s No. 12 seed, but are 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.
Kyrie Irving clearly deserved the award over Holiday
The week covered four Nets games, one in which Kyrie did not play. He scored 15, 30, and 48. Yesterday’s 38 point game counts for this week. I think Holiday was the right choice.
How was his defense? Legitimately asking, because I watched all of Jrue’s games and he doesn’t let up.