CODEWITHSUNDEEP
python
user1819717
user1819717

Reputation: 133

error more than 1 value to unpack

Im not even sure what exactly my problem is from this error. Any information would be very helpful.

what i have so far: 

def equations(specie,elements):
vectors=[]
for x in specie: 
    vector=extracting_columns(x,elements)
    vectors.append(vector)

When i run:

 equations(['OH', 'CO2','c3o3','H2O3','CO','C3H1'], 
 ['H', 'C', 'O'])

i get the following error:

    Traceback (most recent call last):
File "<stdin>", line 1, in <module>

File "_sage_input_77.py", line 10, in exec compile(u'print support.syseval(python, u"equations([\'OH\', \'CO2\',\'c3o3\',\'H2O3\',\'CO\',\'C3H1\'], unel)", SAGE_TMP_DIR) File "", line 1, in

File "/sagenb/sage_install/sage-5.4-sage.math.washington.edu-x86_64-Linux/devel/sagenb-git/sagenb/misc/support.py", line 479, in syseval return system.eval(cmd, sage_globals, locals = sage_globals) File "/sagenb/sage_install/sage-5.4-sage.math.washington.edu-x86_64-Linux/local/lib/python2.7/site-packages/sage/misc/python.py", line 56, in eval eval(z, globals) File "", line 1, in

File "", line 4, in equations

File "", line 3, in extracting_columns

ValueError: need more than 1 value to unpack

my previous functions if needed: import re def parse_formula(formula): '''Given a simple chemical formula, return a list of (element, multiplicity) tuples.

Example:
'H2SO4' --> [('H', 2.0), ('S', 1.0), ('O', 4.0)]


'''

return [ (elem, float(mul) if mul else 1.) for (elem, mul) in re.findall(r'([A-Z][a-z]*)(\d*)', formula) ]

def unique_element(group): c=[] for element in group: piece=parse_formula(element) for x in piece: c.append(x[0])

return list(set(c))

def extracting_columns(specie, elements): species_vector=zeros(len(elements)) for (el,mul) in specie: species_vector[elements.index(el)]=mul

return species_vector

Upvotes: 1

Views: 313

Answers (1)

abarnert
abarnert

Reputation: 365657

The problem is that you're calling extracting_columns with a string like 'OH' as the first argument, so when you try to do for (el,mul) in specie: it's trying to unpack the 'O' into (el, mul).

An easy way to debug this is to insert a print right before the offending line:

def extracting_columns(specie, elements):
  species_vector=zeros(len(elements))
  print(specie)
  for (el,mul) in specie:
    species_vector[elements.index(el)]=mul
  return species_vector

So, how is extracting_columns getting 'OH'? Well, let's look at where it's called, with a couple more prints:

def equations(specie,elements):
  vectors=[]
  print(specie)
  for x in specie:
    print(x)
    vector=extracting_columns(x,elements)
    vectors.append(vector)

Now, when you run it, you'll see that specie is ['OH', 'CO2', 'c3o3', 'H2O3', 'CO', 'C3H1'], so the first element of it is obviously 'OH'.

As for how to fix this… well, without knowing what you're actually trying to do, it's hard to tell you how to do it. But obviously if you want to iterate over the first argument to extracting_columns and treat each item as a pair, you have to pass it a sequence of pairs rather than a string.

But it looks like your parse_formula is made specifically for turning a string like 'OH' into a list of pairs like [('O', 1.0), ('H', 1.0)]. So presumably the problem is that you just forgot to call it somewhere. Maybe you want equations to look like this?

def equations(specie, elements):
  vectors=[]
  for x in specie:
    formula = parse_formula(x)
    vector=extracting_columns(formula, elements)
    vectors.append(vector)

This does something, without any exceptions. Whether it's what you actually want, I have no idea.

At any rate, learning how to see what's actually going on in your code and debug trivial problems is probably more important than getting the right answer here immediately.

Upvotes: 2

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