currently I'm displaying keys only, each in new line:
'<br/>'.join(mydict)
how do I display them like key:: value, each in the new line?
Add a comment
|
5 Answers
Go through the dict.items() iterator that will yield a key, value tuple:
'<br/>'.join(['%s:: %s' % (key, value) for (key, value) in d.items()])
Updated with modern f-string notation:
'<br/>'.join([f'{key}:: {value}' for key, value in d.items()])
1 Comment
Greg Bray
Less verbose if you don't unpack the tuple:
'<br/>'.join(['%s:: %s' % kv for kv in d.items()])That, or an even cooler solution using join to join both items and the (key,value) pairs, avoiding the now-obsolete % interpolation, using only a single dummy variable _, and dropping the redundant list comprehension:
"<br/>".join(":: ".join(_) for _ in mydict.items())
Be aware that dicts have no ordering, so without sorted() you might not get what you want:
>>> mydict = dict(a="A", b="B", c="C")
>>> ", ".join("=".join(_) for _ in mydict.items())
'a=A, c=C, b=B'
This also only work when all keys and values are strings, otherwise join will complain. A more robust solution would be:
", ".join("=".join((str(k),str(v))) for k,v in mydict.items())
Now it works great, even for keys and values of mixed types:
>>> mydict = {'1':1, 2:'2', 3:3}
>>> ", ".join("=".join((str(k),str(v))) for k,v in mydict.items())
'2=2, 3=3, 1=1'
Of course, for mixed types a plain sorted() will not work as expected. Use it only if you know all keys are strings (or all numeric, etc). In the former case, you can drop the first str() too:
>>> mydict = dict(a=1, b=2, c=2)
>>> ", ".join("=".join((k,str(v))) for k,v in sorted(mydict.items()))
'a=1, b=2, c=3'
4 Comments
Jihye Seo
The last one is same data format of my dictionary. But when I try you solution It returns this message : TypeError: 'str' object is not callable
MestreLion
@JihyeSeo: which one of my solutions are you trying, and what is your dict layout? I gave 3 alternatives: the simplest one requires both keys and values to be strings; the more robust one works with mixed data but can't be
sorted(), and the last one expects uniform data (and string keys). It all depends on your data format.MestreLion
@JihyeSeo: Given your error, quite possibly you named a variable
str, shadowing the builtin str class. Don't use builtin names to name your objects!Jihye Seo
The last one I used. I didn't use builtin str class. Anyway I solved it using different solution. Thanks a lot!
In python 3.6 I prefer this syntax, which makes the code even more readable:
'<br/>'.join([f'{key}: {value}' for key, value in d.items()])
Comments
If you wanted to be more pythonic, we can remove the for-comprehension altogether.
'<br/>'.join(map(':: '.join, mydict.items()))
1 Comment
luisfelipe18
just 1
) is missing in the end: '<br/>'.join(map(':: '.join, mydict.items()))I liked what Filip Dupanović did above. I modified this to be python3-ish using str.format method.
'<br/>'.join(['{0}:: {1}'.format(key, value) for (key, value) in d.items()])
I modified the above to generate a string I could feed to a SQL table create statement.
fieldpairs = ','.join(['{0} {1}'.format(key, value) for (key, value) in HiveTypes.items()])
