需要使用最后一个元素对给定的元组列表进行排序
def sort_last(tuples):
sorted_list = []
i = 0
i2 = 1
while True:
if len(tuples) == 2:
if tuples[i][-1] < tuples[i2][-1]:
sorted_list.append(tuples[i])
sorted_list.append(tuples[i2])
break
else:
sorted_list.append(tuples[i2])
sorted_list.append(tuples[i])
break
elif tuples[i][-1] < tuples[i2][-1]:
i2 += 1
print 1
elif i == i2:
i2 += 1
elif i2 == len(tuples)-1:
if tuples[i][-1] < tuples[i2][-1]:
sorted_list.append(tuples[i])
del tuples[i]
i = 0
i2 = 1
else:
sorted_list.append(tuples[i2])
del tuples[i2]
i = 0
i2 = 1
elif len(tuples) <= 1:
return tuples
else:
i += 1
return sorted_list
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使用元组列表,([(1, 3), (3, 2), (2, 1)])或者([(2, 3), (1, 2), (3, 1)])它正确返回:([(2, 1), (3, 2), (1, 3)])和([(3, 1), (1, 2), (2, 3)]) 但它写入 3+ 元素元组或 4 个元组列表"list index out of range"
def sort_last(tuples):
return sorted(tuples, key=lambda i: i[-1])
>>> sort_last([(1, 3), (3, 2), (2, 1)])
[(2, 1), (3, 2), (1, 3)]
>>> sort_last([(2, 3), (1, 2), (3, 1)])
[(3, 1), (1, 2), (2, 3)]
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