As I already asked on (Convert dictionary key tuple to string), but this time I have a dict={("name1","name2","name3"):256}
I want my dict to be dict={"name1,name2,name3":256} but when I try this I get an error:
new_dct = {}
for k,v in dicts.items():
new_dct[','.join(k)] = v
**TypeError: sequence item 1: expected str instance, float found**
(i removed all nanS) Any idea?
Try this:
d = {("name1", "name2", "name3"): 256}
new_d = {','.join(map(str, k)): v for k, v in d.items()}
print(new_d) # {'name1,name2,name3': 256}
I tried your code with dict renamed as d and used in the loop in place of dicts (the value of which is unclear), and it seems to work fine:
d={("name1","name2","name3"):256}
new_dct = {}
for k,v in d.items():
new_dct[','.join(k)] = v
print(new_dct)
Output:
{'name1,name2,name3': 256}
Here are a few notes:
dicts without showing what value you assigned to it.dicts using dicts={("name1", 2.0, "name3"):256}, it would raise the exception you have mentioned (TypeError: sequence item 1: expected str instance, float found).str, otherwise join() may not work.
d in place of the variable dicts with unknown value (and also without using the python type name dict as a variable name).It's the same like before in Convert dictionary key tuple to string.
Just remove 's' in dicts.items() as in your first dict variable or find another typo in your code espescially in element of dict or dicts. The exception (TypeError) could be cause by different reference to your dict or dicts variable. Others' answers already solved it, whether using dict comprehension:
dict = {("name1","name2","name3"):256}
new_dct = {','.join(key): value for key, value in dict.items()}
print(new_dct)
or normal for loop:
dict = {("name1","name2","name3"):256}
new_dct = {}
for key, value in dict.items():
new_dct[','.join(key)] = value
print(new_dct)
and the result will be same output.
{'name1,name2,name3': 256}
dicts={(1.0, "name1"):256}it would raise that exception (plus or minus the item number), and OP has not shown howdictsgets assigned.dict = {("name1", "name2", "name3"): 256}which is their provided sample input apparently and with it, that error gets raised, that's false, with that input and that code, that error can't be raised, you basically said the same thing in your answer (questionable why it's even an answer, but sure)dicts(notdict), and we're not shown what's indicts.dict, because the code references the variabledictswhose value we are not shown, the exception shown in OP's question is entirely possible.