Dataframe Constructor Not Properly Called! Error
I Am New to Python and I Am Facing Problem in Creating the Dataframe in the Format of Key and Value I. E. Data = [{'Key':'\[Globalprogramsizeinthousands\]','Val...
I am new to Python and I am facing problem in creating the Dataframe in the format of key and value i.e.
data = [{'key':'\[GlobalProgramSizeInThousands\]','value':'1000'},]
Here is my code:
columnsss = ['key','value'];
query = "select * from bparst_tags where tag_type = 1 ";
result = database.cursor(db.cursors.DictCursor);
result.execute(query);
result_set = result.fetchall();
data = "[";
for row in result_set:
`row["tag_expression"]`)
data += "{'value': %s , 'key': %s }," % ( `row["tag_expression"]`, `row["tag_name"]` )
data += "]" ;
df = DataFrame(data , columns=columnsss);
But when I pass the data in DataFrame it shows me
pandas.core.common.PandasError: DataFrame constructor not properly called!
while if I print the data and assign the same value to data variable then it works.
2 Answers
You are providing a string representation of a dict to the DataFrame constructor, and not a dict itself. So this is the reason you get that error.
So if you want to use your code, you could do:
df = DataFrame(eval(data))
But better would be to not create the string in the first place, but directly putting it in a dict. Something roughly like:
data = []
for row in result_set:
data.append({'value': row["tag_expression"], 'key': row["tag_name"]})
But probably even this is not needed, as depending on what is exactly in your result_set you could probably:
- provide this directly to a DataFrame:
DataFrame(result_set) - or use the pandas
read_sql_queryfunction to do this for you (see docs on this)
Just ran into the same error, but the above answer could not help me.
My code worked fine on my computer which was like this:
test_dict = {'x': '123', 'y': '456', 'z': '456'}
df=pd.DataFrame(test_dict.items(),columns=['col1','col2'])
However, it did not work on another platform. It gave me the same error as mentioned in the original question. I tried below code by simply adding the list() around the dictionary items, and it worked smoothly after:
df=pd.DataFrame(list(test_dict.items()),columns=['col1','col2'])
Hopefully, this answer can help whoever ran into a similar situation like me.
Sarah Jenkins is a veteran tech journalist with over 12 years of experience covering artificial intelligence, mobile innovations, and digital ethics. Her insights have appeared in leading technology publications worldwide.
