Question

给出一本字典

@IBAction func startButton2(sender: AnyObject) {
    paintCell2(0)
}

func paintCell2(index: Int) {
    let indexPath = NSIndexPath(forItem: index, inSection: 0)
    let cell = self.collectionView.cellForItemAtIndexPath(indexPath)
    cell?.contentView.backgroundColor = UIColor.blackColor()

    if index < 25 {
        let delay = Int64(NSEC_PER_SEC)
        let time = dispatch_time(DISPATCH_TIME_NOW, delay)
        dispatch_after(time, dispatch_get_main_queue()) {
            self.paintCell2(index + 1)
        }
    }
}

我想要一个打印出来的功能：

  x = {'b': {d}, 'a': {'b', 'c'}, 'd': {'g'}, 'f': {'d', 'g'}, 'c': {'e', 'f'}, 'e': {d}}

我写了这个：

  a -> ['b', 'c']
  b -> ['d']
  c -> ['e', 'f']
  d -> ['g']
  e -> ['d']
  f -> ['d', 'g']

它有效，但很糟糕。我怎样才能做得更好？

Answer 1

假设你的意思是：

x = {'b': {'d'}, 'a': {'b', 'c'}, 'd': {'g'}, 'f': {'d', 'g'}, 'c': {'e', 'f'}, 'e': {'d'}}

（与键'b'和'e'关联的值在引号中，与键'a'关联的值为{'b'，'c'}，而不是{'c'，'d'} ）

然后这将给出您列出的输出：

for k in sorted(x.keys()):
    print("{} -> {}".format(k, sorted(list(x[k]))))

Answer 2

x = {'b': {'d'}, 'a': {'c', 'd'}, 'd': {'g'}, 'f': {'d', 'g'}, 'c': {'e', 'f'}, 'e': {'d'}}


def graph_as_str(x):
    sorted_x = sorted(x)

    for i in x:
        print i, "-->", list(x[i])

希望这能解决您的需求，并且更加简单。享受!!!!

Answer 3

假设d也被引用：

x = {'b': {'d'}, 'a': {'c', 'b'}, 'd': {'g'}, 'f': {'d', 'g'}, 
    'c': {'e', 'f'}, 'e': {'d'}}

x_str = '\n'.join([key + ' -> ' + str(sorted(list(val))) 
                   for key, val in sorted(x.iteritems())])
print(x_str)

Answer 4

OP确实请求了一个函数，并给出了一个精确的格式：

x = {'b': {'d'}, 'a': {'c', 'b'}, 'd': {'g'}, 'f': {'d', 'g'}, 'c': {'e', 'f'}, 'e': {'d'}}

def graph_as_str(graph: {str:{str}}) -> str:
    return "\n".join("  {} -> {}".format(key, sorted(value)) for key, value in sorted(graph.items()))

将字典格式化为具有可变数值的字符串

4 个答案: