作业帮python英文题目求解答!~~~~Given dictionaries, d1 and d2, create a new dictionary with the following property: for each entry (a, b) in d1, if a is not a key of d2 (i.e., not a in d2) then add (a,b) to the new dictionary for each entry (a, b) in
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python英文题目求解答!~~~~Given dictionaries, d1 and d2, create a new dictionary with the following property: for each entry (a, b) in d1, if a is not a key of d2 (i.e., not a in d2) then add (a,b) to the new dictionary for each entry (a, b) in
python英文题目求解答!~~~~
Given dictionaries, d1 and d2, create a new dictionary with the following property: for each entry (a, b) in d1, if a is not a key of d2 (i.e., not a in d2) then add (a,b) to the new dictionary for each entry (a, b) in d2, if a is not a key of d1 (i.e., not a in d1) then add (a,b) to the new dictionary
For example, if d1 is {2:3, 8:19, 6:4, 5:12} and d2 is {2:5, 4:3, 3:9}, then the new dictionary should be {8:19, 6:4, 5:12, 4:3, 3:9}
Associate the new dictionary with the variable d3
python英文题目求解答!~~~~Given dictionaries, d1 and d2, create a new dictionary with the following property: for each entry (a, b) in d1, if a is not a key of d2 (i.e., not a in d2) then add (a,b) to the new dictionary for each entry (a, b) in
def new_dict(dict1, dict2):
newdict = {}
newdict.update(dict1)
newdict.update(dict2)
d = newdict.copy()
for i in d.iterkeys():
if dict1.has_key(i) and dict2.has_key(i):
newdict.pop(i)
return newdict