Print value from a list (The truth value of a DataFrame is ambiguous error)

Question

Having a question. I am having a list of records and there is another list of records which i am comparing the first list. When i write line(inside row reading of first list:

    for index, row in output_merged_po.iterrows():
        stock = output_merged_stock[output_merged_stock['PN_STRIPPED']==row['PN_STRIPPED']][['Whs']]
        print stock

I get result

       Whs
 11763 VLN

Where 11763 is output_merged_stock id number and Whs is name of whs where PN_stripped matches.

But i fail to extract data for further processing. I just want to write simple if statetement where i can ask if whs = VLN. I wrote:

                    if stock[['Whs']] == 'VLN':
                         print stock

I got error: The truth value of a DataFrame is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

I wrote:

                    if stock == 'VLN':
                        print stock

And i got again : The truth value of a DataFrame is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

How should i write if statement if i want to get result 'VLN'? As for example there are sometimes cases when stock output is sometimes 3 whs, where 2 of them are 'VLN' and third is 'XRS' and on that case i should see of 'if' output 2 times VLN without XRS

Answer 1

You're trying to compare a df, which is unnecessary here by the way, with a scalar value which is incorrect as it becomes ambiguous for testing a scalar value against because you may have 1 or more matches.

I think you want:

if all(stock['Whs']] == 'VLN'):

or if you know there is only a single row then:

if stock['Whs'].values[0] == 'VLN':

example:

In [79]:
# create some dummy data
df = pd.DataFrame({'a':np.arange(5), 'b':2})
df
Out[79]:
   a  b
0  0  2
1  1  2
2  2  2
3  3  2
4  4  2

try something like what you tried:

if df['a'] == 2:
    print("we have 2")

which raises:

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

So we could take the hint from the error:

In [82]:

if any(df['a'] == 2):
    print("we have 2")
we have 2

We can use all with 'b' column:

In [83]:

if all(df['b'] == 2):
    print("all are 2")
all are 2

If you compared a series which had a single row value then you could do this:

In [84]:

if df.iloc[2]['a'] == 2:
    print("we have 2")
​
we have 2

but it becomes ambiguous with more than 1 row:

if df.iloc[1:3]['b'] == 2:
    print("we have 2")

again raises ValueError but the following would work:

In [87]:

if df.iloc[1:3]['b'].values[0] == 2:
    print("we have 2")
​
we have 2