error using boolean indexing with pandas

Question

I have a column "APNT_NA_ACTN" that provides the type of coding used to hire employees:

115, 515, 100, 786, 101, etc...

I have aliased my set of data as names, therefore names[:3] provides three rows of the entire set.

I have the ability to filter one type of code:

names[names['APNT_NA_ACTN'] == 115]
names

but, I want to filter only: 115 and 515 from this column. I've tried the following

temp = names[(names['APNT_NA_ACTN'] == 115) & (names['APNT_NA_ACTN'] == 515)]
temp

and I have also tried:

temp = names.query('[100,515] in 'APNT_NA_ACTN')

can anyone offer assistance?

thanks

so in all both suggestions below worked for me:

1) temp = names[names['APNT_NA_ACTN'].isin([115,515])]

2) hiring_code = names['APNT_NA_ACTN'] temp = names[(hiring_code == 115) | (hiring_code == 515)] temp[['NM_EMP_LST','NAT_ACTN_2_3','ACTN_YMD','ORG_LEV2','ORG_LEV3','APNT_NA_ACTN','APNT_YMD','SCD_LV_YMD','SSNO','year']]

I like this concept, however I receive the error: isin() takes exactly 2 arguments (3 given) I entered the code like this: names['APNT_NA_ACTN'].isin([115,515)] names — Dave
– Dave, Commented Sep 11, 2014 at 15:33
Sorry typo error, try this: names['APNT_NA_ACTN'].isin([115,515])] — EdChum
– EdChum, Commented Sep 11, 2014 at 15:35

EdChum · Accepted Answer · 2014-09-11 15:37:25Z

2

An alternative is to use isin:

names['APNT_NA_ACTN'].isin([115,515])]

You can pass a list or a Series to the method

answered Sep 11, 2014 at 15:37

EdChum

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Comments

unutbu · Accepted Answer · 2014-09-11 13:11:31Z

1

Use | (logical-or) instead of & (logical-and):

hiring_code = names['APNT_NA_ACTN']
temp = names[(hiring_code == 115) | (hiring_code == 515)]

answered Sep 11, 2014 at 13:11

unutbu

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1 Comment

Dave Over a year ago

this seemed to provide what I needed. totally grateful all input!

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