按第一项对嵌套列表进行排序-itemgetter不能胜任
问题内容：
我有一本字典，我已将其转换为列表，因此可以按第一项进行排序。字典中的键是一个字符串（数字），值是一个保留在列表中的整数。
字典转换的列表如下所示：
[('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), ('89378', 1), ('89379', 1),.........]
字典中大约有240,000个项目。我想按第一个索引对字典进行排序，但是当我使用itemgetter（0）时，它将首先按所有“
1”对列表进行排序。列出的排序如下：
[('0', 3), ('1', 3), ('10', 3), ('100', 4), ('1000', 3), ('10000', 1), ('100000', 3), ('100001', 2), ('100002', 3), ('100003', 3), ('100004', 2), ('100005', 2), ('100006', 2), ('100007', 2), ('100008', 2), ('100009', 2), ('10001', 1), ('100010', 3), ('100011', 3), ('100012', 3), ('100013', 2), ('100014', 1), ('100015', 1), ('100016', 1), ('100017', 1), ('100018', 1), ....]
我希望列表按[‘0’，3），（‘1’，3），（‘2’，整数），（‘3’，整数），…（‘240,000’，整数）]
这是我在其中将文本文件读入字典，转换为列表并使用itemgetter按嵌套列表中的第一项排序的代码。我需要代码中的字典，因为我严重依赖它来通过键查找值。一旦所有进程都运行，我只想对输出文件的字典进行排序。谢谢你的帮助。
import sys, string, csv, arcpy, os, fileinput, traceback
from arcpy import env
from operator import itemgetter
#Creating a dictionary of FID: LU_Codes from external txt file
text_file = open("H:\SWAT\NC\FID_Whole_Copy.txt", "rb")
#Lines = text_file.readlines()
FID_GC_dict = dict()
reader = csv.reader(text_file, delimiter='\t')
for line in reader:
FID_GC_dict[line[0]] = int(line[1])
text_file.close()
dict_List = [(x, FID_GC_dict[x]) for x in FID_GC_dict.keys()]
dict_List.sort(key=itemgetter(0))
print dict_List
问题答案：
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()]
