更改键以将字符串转换为int会有所帮助，这也是其他一些排序技巧。
from operator import itemgetter
list_to_sort=[('89372', 2), ('89373', 1), ('89374', 1), ('89375', 1), ('89376', 1), ('89377', 1), ('228055', 1), ('228054', 1), ('228057', 2), ('228056', 1), ('228051', 1), ('228050', 1),('228053', 1), ('203184', 6), ('228059', 1), ('228058', 1), ('89370', 2), ('89371', 3), ('89372', 2), ('89373', 1), ('89374', 1), ('89375', 1), ('89376', 1), ('89377', 1)]
print list_to_sort
list_to_sort.sort()
print list_to_sort # badly sorted as described
list_to_sort.sort(key=itemgetter(0))
print list_to_sort # badly sorted as described (same as above)
list_to_sort.sort(key=lambda x: int(x[0]))
print list_to_sort # sorted well
list_to_sort.sort(key=lambda x: int(x[0]), reverse=True)
print list_to_sort # sorted well in reverse
关于构建要从字典排序的列表的旁注。iteritems()是执行以下操作的更好方法
dict_List = [(x, FID_GC_dict[x]) for x in FID_GC_dict.keys()]
dict_List = [(k,v) for k,v in FID_GC_dict.iteritems()]
602392714
清零编程群