Hi guys, since its so quiet here

Question

Hi guys, since its so quiet here

kingz2000 (Programmer)

(OP)

11 May 20 09:11

Hi guys,

since its so quiet here ,let me throw in another easy one. The problem is, that I'm so used toSQL and using "Where"..So here goes..

df['artist'].value_counts().reset_index()

gives me a select count...

now how I can add a condition to only list those above a certain number, say 3?

I tried this...

df.groupby('artist').count()>3

..which just gave me my list with true and false on each column.

Thanks in advance

xwb (Programmer) · Answer 1 · 2020-05-11 10:04:20

RE: Hi guys, since its so quiet here

xwb (Programmer) 11 May 20 10:04

Is df a panda's dataframe?

kingz2000 (Programmer) · Answer 2 · 2020-05-11 10:37:31

yes, df is the general short term for any panda dataframe

mikrom (Programmer) · Answer 3 · 2020-05-11 13:16:40

kingz2000,
it would be helpful if you could post how your data looks like.

mikrom (Programmer) · Answer 4 · 2020-05-11 20:41:13

I tried this:
first I created this data frame from a list of items

CODE

>>> my_list = ['foo', 'bar', 'baz', 'foo', 'bar', 'foo', 'bar', 'foo']

>>> my_df=pd.DataFrame({"item" : my_list})

>>> my_df
  item
0  foo
1  bar
2  baz
3  foo
4  bar
5  foo
6  bar
7  foo

Now I can get a list of items which occurs more than once in these steps:

CODE

>>> my_df["item"].value_counts()>1
foo     True
bar     True
baz    False
Name: item, dtype: bool

>>> my_df.loc[(my_df["item"].value_counts()>1).values]
  item
0  foo
1  bar

>>> list((my_df.loc[(my_df["item"].value_counts()>1).values])["item"])
['foo', 'bar']

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Hi guys, since its so quiet here