CODEWITHSUNDEEP
pythonpython-2.7pandas
Jason
Jason

Reputation: 4546

Unexpected NameError occurs when trying to use a dataframe name defined within a function

Could somebody explain why the following code produces a NameError?

def nonull(df, col, name):
    name = df[pd.notnull(df[col])]
    print name[col].count(), df[col].count()
    return name

nonull(sve, 'DOC_mg/L', 'sveDOC')
sveDOC.count()

NameError: name 'sveDOC' is not defined

711 711

The dataframe seems to be created as the print statement works, so I don't understand why when I try to use sveDOC (which was name inside the function) it produces an error.

Here's an example of what I'd like to do within the function:

import pandas as pd

d = {'one' : pd.Series([1., 1., 1., 1.], index=['a', 'b', 'c', 'd']), 
     'two' : pd.Series([1., 2., 3., 4.], index=['a', 'b', 'c', 'd'])}
pd.DataFrame(d)
df = pd.DataFrame(d)
df1 = df
df = df * 2
print df.head(), df1.head()

one  two
a    2    2
b    2    4
c    2    6
d    2    8    
one  two
a    1    1
b    1    2
c    1    3
d    1    4

Upvotes: 1

Views: 7193

Answers (1)

jonrsharpe
jonrsharpe

Reputation: 122107

Python names do not work the way you seem to think. Here's what your code actually does:

def nonull(df, col, name):
    name = df # rebind the name 'name' to the object referenced by 'df'
    name = df[pd.notnull(name[col])] # rebind the name 'name' again 
    print name[col].count(), df[col].count()
    return name # return the instance

nonull(sve, 'DOC_mg/L', 'sveDOC') # call the function and ignore the return value

The function never actually uses the 'sveDOC' argument. Here's what you should actually do:

def nonull(df, col):
    name = df[pd.notnull(df[col])]
    print name[col].count(), df[col].count()
    return name

sveDOC = nonull(sve, 'DOC_mg/L')
sveDOC.count()

Your conception of Python's use of names and references is completely wrong.

pd.DataFrame(d) # creates a new DataFrame but doesn't do anything with it
                # (what was the point of this line?)
df = pd.DataFrame(d) # assigns a second new DataFrame to the name 'df'
df1 = df # assigns the name `df1` to the same object that 'df' refers to
         # - note that this does *not* create a copy
df = df * 2 # create a new DataFrame based on the one referenced by 'df' 
            # (and 'df1'!)and assign to the name 'df'

To demonstrate this:

df1 = pd.DataFrame(d)

df2 = df1

df1 is df2
Out[5]: True # still the same object

df2 = df2 * 2

df1 is df2
Out[7]: False # now different

If you want to create a copy of a DataFrame, do so explicitly:

df2 = copy(df1)

You can either do this outside nonull and pass the copy, or do it inside nonull and return the modified copy.

Upvotes: 2

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