If I use
df.loc[(origFull['abc'] == 111)]
it will work of course.
Then
test = '(origFull[\'abc\'] == 111)'
df.loc[@test]
or
df.loc['@test']
Neither worked.
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3 Answers
A few things are wrong:
1: You want test to be a value(a boolean value to be exact), if you added single quotes around them, test turns into a string. Do this instead:
test = origFull['abc'] == 111
2: @ symbol isn't valid syntax for python, excepted when it's used for decorators. To use test as a variable, just write test without the @.
df.loc[test]
1 Comment
Windtalker
actually I want test variable to represent a statement, but you are right, I do not need to add single quote. Thank you!
Take the quotes away when you define test.
Inside of loc you are essentially creating a boolean variable. So set test equal to that boolean and use it as a boolean index:
test = (origFull['abc'] == 111)
df.loc[test]
2 Comments
Windtalker
It will work after removing the back slash. Thank you very much, since you are the first one answered this question and brought me to the right direction!
Reen
Yes, sorry about that, just copied and pasted and missed the backslashes.
Maybe you are looking for the .query() method?
test = "abc == 111"
df.query(test)
1 Comment
Windtalker
in fact it's .loc, but I got the answer. thanks anyway