How to pass a dictionary as a function value in python

Question

In python, I use mincemeat chart reduction infrastructure

From my map function, I would like yield (k,v) in a loop that would send the result of the decrease function (sample data that are the result of my display function)

 auth3 {'practical': 1, 'volume': 1, 'physics': 1} auth34 {'practical': 1, 'volume': 1, 'chemistry': 1} .... 

There would be many such records; these are just a few examples.

Here auth3 and auth34 are keys, and the corresponding values ​​are dictionary elements

Inside the reduction function, when I try to print the key values, I get the error "too many values ​​to decompress". My reduction function looks like this:

 def reducefn(k, v): for k,val in (k,v): print k, v 

Please let me know how to resolve this error.

Answer 1

Define your dictionary first with the built-in python dict

 >>> dic1 = dict(auth3 = {'practical': 1, 'volume': 1, 'physics': 1}, auth34 = {'practical': 1, 'volume': 1, 'chemistry': 1} ) >>> dic1 {'auth3': {'practical': 1, 'volume': 1, 'physics': 1}, 'auth34': {'practical': 1, 'volume': 1, 'chemistry': 1}} 

Then your reduction function can go like

 def reducefn(dictofdicts): for key, value in dictofdicts.iteritems() : print key, value 

Finally,

 >>> reducefn(dic1) auth3 {'practical': 1, 'volume': 1, 'physics': 1} auth34 {'practical': 1, 'volume': 1, 'chemistry': 1} 

Answer 2

 def reducefn(*dicts): #collects multiple arguments and stores in dicts for dic in dicts: #go over each dictionary passed in for k,v in dic.items(): #go over key,value pairs in the dic print(k,v) reducefn({'practical': 1, 'volume': 1, 'physics': 1} ,{'practical': 1, 'volume': 1, 'chemistry': 1}) 

Gives out

 >>> physics 1 practical 1 volume 1 chemistry 1 practical 1 volume 1 

Now about your implementation:

 def reducefn(k, v): 

The above function has two arguments. The arguments passed to the function are accessed via k and v respectively. Thus, calling reducefn({"key1":"value"},{"key2":"value"}) causes k assigned {"key1":"value"} and v assigned {"key2":"vlaue"} .

When you try to call it like this: reducefn(dic1,dic2,dic3,...) you pass more than the allowed number of parameters, as defined by the reducefn declaration / signature.

for k,val in (k,v):

Now, assuming that you went through two dictionaries to reducefn , both k and v will be dictionaries. The for loop above would be equivalent:

 >>> a = {"Name":"A"} >>> b = {"Name":"B"} >>> for (d1,d2) in (a,b): print(d1,d2) 

Which gives the following error:

 ValueError: need more than 1 value to unpack 

This is because you do this when the for loop is called:

 d1,d2=a 

You can see that we get this error when we try to do it in REPL

 >>> d1,d2=a Traceback (most recent call last): File "<pyshell#24>", line 1, in <module> d1,d2=a ValueError: need more than 1 value to unpack 

We could do this:

 >>> for (d1,d2) in [(a,b)]: print(d1,d2) {'Name': 'A'} {'Name': 'B'} 

Assigns a tuple (a,b) to d1,d2 . This is called unpacking and will look like this:

d1,d2 = (a,b)

However, in our loop for k,val in (k,v): that would not make sense, since we would end up with k , and val representing the same thing as k , v . Instead, we need to go through the key value pairs in the dictionaries. But, seeing that we need to cope with n dictionaries, we need to rethink the definition of a function.

Hence:

 def reducefn(*dicts): 

When you call the function as follows:

 reducefn({'physics': 1},{'volume': 1, 'chemistry': 1},{'chemistry': 1}) 

*dicts collects arguments in such a way that dicts ends as:

 ({'physics': 1}, {'volume': 1, 'chemistry': 1}, {'chemistry': 1}) 

As you can see, the three dictionaries passed to the function were collected in a tuple. Now we iterate over the tuple:

 for dic in dicts: 

So, now, at each iteration, dic is one of the dictionaries we went through, so now we go ahead and print out pairs of keys, values ​​in it:

 for k,v in dic.items(): print(k,v) 

Answer 3

Use zip

 def reducefn(k, v): for k,val in zip(k,v): print k, v >>> reducefn({'practical': 1, 'volume': 1, 'physics': 1} ,{'practical': 1, 'volume': 1, 'chemistry': 1}) practical {'practical': 1, 'volume': 1, 'chemistry': 1} volume {'practical': 1, 'volume': 1, 'chemistry': 1} physics {'practical': 1, 'volume': 1, 'chemistry': 1} >>> 

reducefn(k,v) : is a set of tuples ((k1,k2,k3..), (v1,v2,v3...))

zippping them gives you ((k1,v1), (k2,v2), (k3,v3)...) and what you want