Python：使用for循环水平打印字典内容

Question

I would like to print out the keys of a dictionary horizontal and on every third key to make a new line. I wrote this like that:

fields= {1 : "not", 2 : "not", 3 : "not",
         4 : "not", 5 : "not", 6 : "not",
         7 : "not", 8 : "not", 9 : "not"} 
for i in fields:
print(i)
if i%3==0:
    print ("\n")

However, the output is then just horizontal with two newlines between every third number.

The ouput I would like to have is this:

123

456

789

Answer 1

Is this what you want ?

fields= {1 : "not", 2 : "not", 3 : "not",
         4 : "not", 5 : "not", 6 : "not",
         7 : "not", 8 : "not", 9 : "not"}
for Key,values in fields.items():
    print(Key, end = "")#python3 , if in python2 use print item, instead.
    if Key%3==0:
        print ("\n")


123

456

789

Answer 2

If you use python 3, or import the print function for python 2, you can specify the end parameter of print, like this: print(i, end="") , and there won't be a newline between the dictionary keys. But as dictionary keys are not ordered, they will be printed in a "random" order, to avoid this, make a list of the keys, and sort it before printing, like this: keys = sorted(dict.keys())

Answer 3

print (i), will do the trick for you. On a side note, the indentation of your code is not correct.

fields= {1 : "not", 2 : "not", 3 : "not",
         4 : "not", 5 : "not", 6 : "not",
         7 : "not", 8 : "not", 9 : "not"} 
for i in fields:
  print (i),
  if i%3==0:
    print ("\n")`

Answer 4

Although the solutions work for the given case, I would like to explain something to you. dictionary in python is not ordered. In this case it is fine because of the way you declared it with just simple 1,2,3... keys. You might think that's how dict work. NO !

>>> a = {1 : "not", 2 : "not", 3 : "not",4 : "not", 5 : "not", 6 : "not",7 : "not", 8 : "not", 9 : "not"}
>>> a
{1: 'not', 2: 'not', 3: 'not', 4: 'not', 5: 'not', 6: 'not', 7: 'not', 8: 'not', 9: 'not'}

Good right so in this case all the solutions would work. However if you think the same and try the same code for the following dict

>>> a= {12 : "not", 21 : "not", 30 : "not",4 : "not", 15 : "not", 61 : "not",71 : "not", 8 : "not", 19 : "not"}

The output would be not in the same order.

>>> a
{19: 'not', 4: 'not', 21: 'not', 71: 'not', 8: 'not', 12: 'not', 61: 'not', 30: 'not', 15: 'not'}

And if you try to iterate and print using for keys in a: you would run into many problems.

So you want the desired output (dict is not ordered so unpredictable) use OrderedDict

from collections import OrderedDict
fields = [(1, 'not'), (2, 'not'), (3, 'not'),
          (4, 'not'), (5, 'not'), (6, 'not'),
          (7, 'not'), (8, 'not'), (9, 'not')]
a = OrderedDict(fields)
for i,key in enumerate(a):
    print(key,end='')
    i+=1
    if i%3==0:
        print()

output:

123
456
789

Also enumerate() should be used to print every three values. Don't rely on the key. What if you key changes?

i,key of enumerate(dictionary) just gives out values like these 0,key1 , 1,key2 , 2,key3 ,...and so on. Just gives out pairs of number,item that's all.

Try changing the keys and use this code and other codes and you'll notice how much you change the results will be same for this solution. And not for the others. Remember always your code should be adaptable. Shouldn't entirely depend on only specific sets of input

123
456
789

Also Note just print() would do if you want them separated into newlines. use `print("\\n") if you want a gap (newline) between them.

123

456

789