3 回答

TA貢獻(xiàn)1813條經(jīng)驗(yàn) 獲得超2個(gè)贊
把它變成字典:
mytup = [('a',2),('a',6),('b',4),('a',4),('b',10),('c',4),('c',6),('c',8),('d',12),('d',10)]
d = {}
for key, value in mytup:
if d.get(key) < value: # d.get(key) returns None if the key doesn't exist
d[key] = value # None < float('-inf'), so it'll work
result = d.items()

TA貢獻(xiàn)1895條經(jīng)驗(yàn) 獲得超7個(gè)贊
我認(rèn)為這應(yīng)該工作:
dict = {}
for key, val in mytup:
try:
if dict[key] < val:
dict[key] = val
except IndexError:
dict[key] = val

TA貢獻(xiàn)1856條經(jīng)驗(yàn) 獲得超17個(gè)贊
Itertools是您的朋友,一線解決方案:
from itertools import groupby
print [ max(g) for _, g in groupby(sorted(mytup), lambda x: x[0] )]
結(jié)果:
[('a', 6), ('b', 10), ('c', 8), ('d', 12)]
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