遍历字典值并随后打印答案

【问题标题】：Loop through dictionary values and print subsequently遍历字典值并随后打印
【发布时间】：2018-11-21 09:10:30
【问题描述】：

我正在尝试以层次结构格式打印以下字典

fam_dict{'6081740103':['60817401030000','60817401030100','60817401030200',
'60817401030300','60817401030400','60817401030500','60817401030600']

如图所示：

60817401030000
    60817401030100
        60817401030200
            60817401030400
                60817401030500
                    60817401030600

到目前为止，我有以下代码有效，但我必须在每一行中手动输入第 i 个索引。如何以递归格式重新调整此代码，而不必每次计算代码行数并手动放置索引值

  my_p = node(fam_dict['6081740103'][0], None)
    my_c = node(fam_dict['6081740103'][1], my_p)
    my_d = node(fam_dict['6081740103'][2], my_c)
    my_e = node(fam_dict['6081740103'][4], my_d)
    my_f = node(fam_dict['6081740103'][5], my_e)
    my_g = node(fam_dict['6081740103'][6], my_f)

    print (my_p.name)
    print_children(my_p)

【问题讨论】：

标签： python dictionary for-loop classification hierarchy

【解决方案1】：

你可以试试这个：

fam_dict = {'6081740103':['60817401030000','60817401030100','60817401030200',
'60817401030300','60817401030400','60817401030500','60817401030600']}

for i, val in enumerate(fam_dict['6081740103']):
    print(' ' * i * 4 + val)

输出你想要的层次结构：

60817401030000
    60817401030100
        60817401030200
            60817401030300
                60817401030400
                    60817401030500
                        60817401030600

【讨论】：

【解决方案2】：

您可以创建一个变量来存储您正在迭代的行，然后在每次循环中递增该变量。您可以将该变量乘以 \t 这是制表符运算符，以控制您想要多少个制表符。这是一个例子：

行 = 0

fam_dict = {'6081740103': ['60817401030000','60817401030100','60817401030200',
           '60817401030300','60817401030400','60817401030500','60817401030600']}

for k, val in fam_dict.items():
   for v in val:
       lines += 1
       t = '\t'
       t = t * lines
       print(t + str(v))

这是你的输出：

60817401030000
    60817401030100
        60817401030200
            60817401030300
                60817401030400
                    60817401030500
                        60817401030600

【讨论】：

【解决方案3】：

你也可以这样做。

for key in fam_dict.keys():
    for i in range(len(fam_dict[key])):
        print(i*"\t"+ fam_dict[key][i])

【讨论】：

【解决方案4】：

这是一个例子：

fam_dict = {'6081740103':['60817401030000','60817401030100','60817401030200','60817401030300','60817401030400','60817401030500','60817401030600']}
for k, v in fam_dict.items():
    for i, s in enumerate(v):
        print("%s%s"% ("\t"*i, s))

如果你想为它制作节点：

fam_dict = {'6081740103':['60817401030000','60817401030100','60817401030200','60817401030300','60817401030400','60817401030500','60817401030600']}
node_list = []
for k, v in fam_dict.items():
    last_parent = none
    for i, s in enumerate(v):
        print("%s%s"% ("\t"*i, s))
        node_list.append(node(v, last_parent))
        last_parent=node_list[-1]

父节点将是node_list[0]。

【讨论】：

【解决方案5】：

试试这个：

fam_dict = {'6081740103':['60817401030000','60817401030100','60817401030200',
'60817401030300','60817401030400','60817401030500','60817401030600']}
l = fam_dict['6081740103']
for i in l:
   print(' '*l.index(i)*4+i)

输出：

60817401030000
    60817401030100
        60817401030200
            60817401030300
                60817401030400
                    60817401030500
                        60817401030600

【讨论】：