I am trying to convert dictionaries into string
for example:
a={3:4,5:6}
s='3 4, 5 6'
The way I am trying is
s=''
i=0
for (k,v) in d.items():
s=s+str(k)+' '+str(v)
while i < len(s):
if s[i]==str(v) and s[i+1]==str(k):
s+=s+s[i]+','+s[i+1]
Here's a Pythonic way of doing that using a list comprehension:
s = ', '.join([str(x) + ' ' + str(a[x]) for x in a])
Output:
'3 4, 5 6'
Update: As Julien Spronck mentioned, the square brackets ([ and ]) are not necessary. Thus, the following has the same effect:
s = ', '.join(str(x) + ' ' + str(a[x]) for x in a)
You can write the list comprehension expression to get the list of key-value string, then join them on , as
>>> d = {3:4,5:6}
>>> ', '.join('{} {}'.format(k, v) for k, v in d.items())
'3 4, 5 6'
OR, even using repr() (Note: it is a hack, I consider .join() approach more pythonic):
>>> repr(d)[1:-1].replace(':', '')
'3 4, 5 6'
I would use the following with a list comprehension:
', '.join([str(k) + ' ' + str(v) for k, v in a.items()])
To illustrate:
In [1]: a = {3:4, 5:6}
In [2]: s = ', '.join([str(k) + ' ' + str(v) for k, v in a.items()])
In [3]: s
Out[3]: '3 4, 5 6'
a[x].
items()?Of course the most pythonic solution is the one using the list comprehension (see @SumnerEvans and the rest) but just for the sake of having an alternative i will post this here:
a = {3: 4, 5: 6}
v = str(a)
for rep in ['{', '}', ':']:
v = v.replace(rep, '')
print(v) # prints -> 3 4, 5 6
Everything can be converted into a string and the manipulated as one. This is what is being done here. From dict to string and then chained replace methods.
timeit now since i am at work.. Feel free to try though.You can still have it like this
s=''
i=0
a={3:4,5:6}
for (k,v) in a.items():
if i < len(a) and s == '':
s = s + str(k) + ' '+ str(v)
elif i < len(a) and s != '':
s += ", " + str(k) + ' '+ str(v)
i += 1
print(s)
This should give you
s='3 4, 5 6'
