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

having question. having list of records , there list of records comparing first list. when write line(inside row reading of first list:

    index, row in output_merged_po.iterrows():         stock = output_merged_stock[output_merged_stock['pn_stripped']==row['pn_stripped']][['whs']]         print stock 

i result

       whs  11763 vln 

where 11763 output_merged_stock id number , whs name of whs pn_stripped matches.

but fail extract data further processing. want write simple if statetement can ask if whs = vln. wrote:

                    if stock[['whs']] == 'vln':                          print stock 

i got error: the truth value of dataframe ambiguous. use a.empty, a.bool(), a.item(), a.any() or a.all().

i wrote:

                    if stock == 'vln':                         print stock 

and got again : the truth value of dataframe ambiguous. use a.empty, a.bool(), a.item(), a.any() or a.all().

how should write if statement if want result 'vln'? example there cases when stock output 3 whs, 2 of them 'vln' , third 'xrs' , on case should see of 'if' output 2 times vln without xrs

you're trying compare df, unnecessary here way, scalar value incorrect becomes ambiguous testing scalar value against because may have 1 or more matches.

i think want:

if all(stock['whs']] == 'vln'): 

or if know there single row then:

if stock['whs'].values[0] == 'vln': 

example:

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

try tried:

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

which raises:

valueerror: truth value of series ambiguous. use a.empty, a.bool(), a.item(), a.any() or a.all().

so take hint error:

in [82]:  if any(df['a'] == 2):     print("we have 2") have 2 

we can use all 'b' column:

in [83]:  if all(df['b'] == 2):     print("all 2") 2 

if compared series had single row value this:

in [84]:  if df.iloc[2]['a'] == 2:     print("we have 2") ​ have 2 

but becomes ambiguous more 1 row:

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

again raises valueerror following work:

in [87]:  if df.iloc[1:3]['b'].values[0] == 2:     print("we have 2") ​ have 2