Question

我有一个如下字典：

d = {
  1: [‘a’,’b’],
  2: [‘c’,’d’],
  8: [‘l’,’p’],
  12: [‘u’,’v’,’w’,’x’]
}

我正在使用iteritems（）迭代字典。如何在迭代时找到字典中每个项目的上一个和下一个项目？

Answer 1

要实现这一点，必须将密钥转换为列表并依赖于获得的订单：

d = {
  1: ['a','b'],
  2: ['c','d'],
  8: ['l','p'],
  12: ['u','v','w','x']
}

def getValBeforeAndAfter(d, the_key):
    # convert keys into the list
    d_list = list(sorted(d))
    index = None
    if (the_key in d_list):
        index = d_list.index(the_key)
        try:
            key_before = d_list[index-1]
            key_after = d_list[index+1]
            return str(key_before) + ": " + str(d[key_before]) + "\n" + str(key_after) + ": " + str(d[key_after])
        except: print "Out of range. You are on the edge of the dictionary"        
    else:
        return "No such key in dictionary"

"""
Small test. Expected output: 
1: ['a', 'b']
8: ['l', 'p']
"""
print getValBeforeAndAfter(d, 2)

Answer 2

from collections import OrderedDict

d2 = {}
d2 = OrderedDict(d2)
d2.update({1: ['a','b']})
d2.update({2: ['c','d']})
d2.update({8: ['l','p']})
d2.update({12: ['u','v','w','x']})

def ajacent(di,key):
    li = list(di.items())
    for i, item in enumerate(li):
        k,v = item
        if k == key and i > 0:
            print(li[i-1])
            print(item)
            print(li[i+1])


print(ajacent(d2,2))

(1, ['a', 'b'])
(2, ['c', 'd'])
(8, ['l', 'p'])

找到每个字典项的下一个和上一个元素

2 个答案: