最简洁的方法是 select 行，其中任何列包含 Pandas dataframe 中的字符串？

Question

What is the most concise way to select all rows where any column contains a string in a Pandas dataframe?

For example, given the following dataframe what is the best way to select those rows where the value in any column contains a b ?

df = pd.DataFrame({
    'x': ['foo', 'foo', 'bar'],
    'y': ['foo', 'foo', 'foo'],
    'z': ['foo', 'baz', 'foo']
})

I'm inexperienced with Pandas and the best I've come up with so far is the rather cumbersome df[df.apply(lambda r: r.str.contains('b').any(), axis=1)] . Is there a simpler solution?

Critically, I want to check for a match in any columns, not a particular column. Other similar questions, as best I can tell, only address a single or list of columns.

Answer 1

This question was not given an answer.. but the question itself and the comments has got the answer already which worked really well for me.. and I didn't find the answer anywhereelse I looked.

So I just copy pasted the answer for someone who can find it useful. I added case=False for a case insensitive serach

Solution from @Reason:

the best I've come up with so far is the rather cumbersome

this one worked for me.

df[df.apply(lambda r: r.str.contains('b', case=False).any(), axis=1)] 

Solution from @rbinnun:

this one worked for me for a test dataset.. but for some real data set.. it returned a unicode error as below, but generally a good solution too I think

df[df.apply(lambda row: row.astype(str).str.contains('b', case=False).any(), axis=1)]

takes care of non-string columns, nans, etc.

UnicodeEncodeError: 'ascii' codec can't encode character u'\xae' in position 5: ordinal not in range(128)

Answer 2

df.apply is too slow when working with lots (millions) of rows. Look for something else.

Answer 3

If you don't like apply:

df.stack()[df.stack().str.contains("b")]

returns

1  z    baz
2  x    bar
dtype: object

and like above with similar to original table properties:

df.stack()[df.stack().str.contains("b")].reset_index().pivot(index="level_0", columns="level_1").droplevel(0, 1)

level_1	x	z
1	NaN	baz
2	bar	NaN