Using loops in python to generate variable

Question

Can I use a loop in Python to generate 10 different variables instead of computing the value of each variable separately? I can imagine doing this in C/C++ wherein I could use an index value to iterate in a loop and generate value.

v1=Levenshtein.jaro_winkler(exhibitor_name,urlparse(search_result1).netloc.encode('utf-8'))

v2=Levenshtein.jaro_winkler(exhibitor_name,urlparse(search_result2).netloc.encode('utf-8'))

v3=Levenshtein.jaro_winkler(exhibitor_name,urlparse(search_result3).netloc.encode('utf-8'))

v4=Levenshtein.jaro_winkler(exhibitor_name,urlparse(search_result4).netloc.encode('utf-8'))

v5=Levenshtein.jaro_winkler(exhibitor_name,urlparse(search_result5).netloc.encode('utf-8'))

v6=Levenshtein.jaro_winkler(exhibitor_name,urlparse(search_result6).netloc.encode('utf-8'))

v7=Levenshtein.jaro_winkler(exhibitor_name,urlparse(search_result7).netloc.encode('utf-8'))

v8=Levenshtein.jaro_winkler(exhibitor_name,urlparse(search_result8).netloc.encode('utf-8'))

v9=Levenshtein.jaro_winkler(exhibitor_name,urlparse(search_result9).netloc.encode('utf-8'))

v10=Levenshtein.jaro_winkler(exhibitor_name,urlparse(search_result10).netloc.encode('utf-8'))

Tim Brown · Accepted Answer · 2014-04-12 18:49:42Z

3

The more pythonic way to do it would be to either do it as a list:

vals = []
for search_result in search_results: 
    vals.append(Levenshtein.jaro_winkler(exhibitor_name,urlparse(search_result).netloc.encode('utf-8')))
# Access via vals[0], vals[1], etc.

or as a dictionary:

vals = {}
for search_result in search_results: 
    vals[search_result] = Levenshtein.jaro_winkler(exhibitor_name,urlparse(search_result).netloc.encode('utf-8'))
# Access via vals[search_result1], vals[search_result2], etc.

If you HAD to be bad about it, you could do:

for i in xrange(10):
    search_result = locals()['search_result' + str(i)]
    locals()['v' + str(i)] = Levenshtein.jaro_winkler(exhibitor_name,urlparse(search_result).netloc.encode('utf-8'))
# Accessed via v0, v1, v2, etc.

But I would recommend against it because it is non-pythonic and more obtuse than the above solutions.

answered Apr 12, 2014 at 18:49

Tim Brown

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1 Comment

DSM Over a year ago

Aside from obtuseness, it's not even guaranteed to work.

Darius Bacon · Accepted Answer · 2014-04-12 18:45:21Z

1

This would be easy if you represented your values as collections. Code structure mirrors data structure; when your data is distributed over unrelated variables, your code is too.

vs = [Levenshtein.jaro_winkler(exhibitor_name, urlparse(search_result).netloc.encode('utf-8'))
      for search_result in search_results]

answered Apr 12, 2014 at 18:45

Darius Bacon

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tmrlvi Over a year ago

@user3450198 if you want to get the variables from namespace you could run

search_results = [locals()[key] for key in locals() if key.startswith("search_result") and not key.startswith("search_results")]

DJG · Accepted Answer · 2014-04-12 18:45:34Z

0

You can use a list like so:

lst = []
for search_result in search_results:
    lst.append(
      Levenshtein.jaro_winkler(
          exhibitor_name,
          urlparse(search_results).netloc.encode('utf-8')))

This assumes that you also used a list, search_results instead of all of the indivdual search_resultN variables.

answered Apr 12, 2014 at 18:45

DJG

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2 Comments

user3450198 Over a year ago

How can I use the above code to generate the variables v1,v2,v3...v10 ?

DJG Over a year ago

@user3450198 See Tim Brown's answer

ely · Accepted Answer · 2014-04-12 20:07:07Z

0

You can use locals() to get a dict that contains the variable names as keys and their contents as values. In this way, you can access the search_result* variables programmatically, instead of typing their names out.

You could then push a new entry into locals() for the v* variable names, but why pollute the local namespace with them? They are all related to each other, so placing them into a dictionary with keys v1 through v10 keeps them together in a useful way.

apply_Levenshtein = lambda x: (  
    Levenshtein.jaro_winkler(exhibitor_name,urlparse(x).netloc.encode('utf-8'))
) 

variables = {
    "v{}".format(i):apply_Levenshtein(locals()["search_result{}".format(i)]) 
    for i in range(1, 11)
}

Now variables["v1"] will be

Levenshtein.jaro_winkler(exhibitor_name,urlparse(search_result1).netloc.encode('utf-8'))`.

Better still would be to ensure that the search_result* variables exist in a dict or a list from the beginning, so that you don't need to use locals() at all.

edited Apr 12, 2014 at 20:07

answered Apr 12, 2014 at 18:47

ely

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1 Comment

DSM Over a year ago

Note that Python isn't under any obligation to pay attention to what you do to what locals() returns, as the docs warn. (PS: vars is an occasionally handy builtin, and so an unfortunate variable name..)

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