Question

我正在使用嵌套列表在我创建的字典中查找值。然后我想将找到的值附加到列表中。我不知道如何编码的问题是如何将值保存在同一嵌套列表结构中？这是我将值附加到空列表的最后一行的代码。

#Creating a dictionary of FID: LU_Codes from external txt file
import sys, arcpy, string, csv

text_file = open("H:\SWAT\NC\FID_Whole.txt", "r")
Lines = text_file.readlines()
text_file.close()

FID_LU = map(string.split, Lines)
#print FID_LU
FID_GC_dict = dict(FID_LU)

Neighbors_file = open("H:\SWAT\NC\Sh_Neighbors2.txt","r")
Entries = Neighbors_file.readlines()
Neighbors_file.close()

Neighbors_List = map(string.split, Entries)

print Neighbors_List

#FID = [x[0] for x in Neighbors_List]
#print FID

gridList = []
for list in Neighbors_List:
    for item in list:
       #print FID_GC_dict[item]
       gridList.append(int(FID_GC_dict[item]))


 print gridList

这是邻居列表的输出（正确）：

[['0', '1', '11', '12', '13'], ['1', '0', '2', '12', '13', '14'], ['2', '1', '3', '13', '14', '15'], ['3', '2', '4', '14', '15', '16'], ['4', '3', '5', '15', '16', '17'], ['5', '4', '6', '16', '17', '18'], ['6', '5', '7', '17', '18', '19'], ['7', '6', '8', '18', '19', '20'], ['8', '7', '9', '19', '20', '21'], ['9', '8', '20', '21', '22'], ['10', '11']]

这是gridList的输出（不正确）：

[3, 3, 4, 4, 4, 3, 3, 3, 4, 4, 4, 3, 3, 3, 4, 4, 2, 3, 3, 3, 4, 2, 2, 3, 3, 3, 2, 2, 2, 3, 3, 3, 2, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4]

我希望gridList看起来像是：

[[3, 3, 4, 4, 4], [3, 3, 3, 4, 4, 4], [3, 3, 3, 4, 4, 2], [3, 3, 3, 4, 2, 2], [3, 3, 3, 2, 2, 2], [3, 3, 3, 2, 2, 3], [3, 3, 3, 2, 3, 3], [3, 3, 3, 3, 3, 3], [3, 3, 3, 3, 3, 3], [3, 3, 3, 3, 3], [3, 4]]

任何帮助将不胜感激。我是python的新手...阅读帖子有帮助，但我正在努力解决这个问题。

谢谢！

Answer 1

制作临时列表row。将内部循环中的项目附加到row，然后在外部循环中将行追加到gridList：

gridList = []
for nlist in Neighbors_List:
    row = []
    for item in nlist:
       row.append(int(FID_GC_dict[item]))
    gridList.append(row)

请注意，您也可以在此处使用list comprehension：

gridList = [[int(FID_GC_dict[item]) for item in nlist] 
            for nlist in Neighbors_List]

PS。最好不要命名变量list，因为它会隐藏同名的内置类型。

Answer 2

您要附加到单个列表中。试试这个

gridList = []
for list in Neighbors_List:
    temp = []
    for item in list:
       #print FID_GC_dict[item]
       temp.append(int(FID_GC_dict[item]))
    gridList.append(temp)

Answer 3

gridList = [int(FID_GC_dict[item]) for item in l for l in Neighbors_List]

Python的list comprehensions非常棒。学习他们，爱他们。（注意这会返回一个元组列表，如果你想要一个列表列表，你可以嵌套理解。

另外，请勿将list用作变量名称，因为它与内置list冲突

Answer 4

使用一个列表：

_list = []; outer_len = 5; inner_len = 10

for outer_idx in range(outer_len):
    _list.append([])
    for inner_idx in range(inner_len):
        _list[outer_idx].append(#stuff)

附加到嵌套列表

4 个答案: