Select by partial string from a pandas DataFrame

re.search(pattern, cell_in_question)

Answer #1:

Based on github issue #620, it looks like you’ll soon be able to do the following:

df[df['A'].str.contains("hello")]

Update: vectorized string methods (i.e., Series.str) are available in pandas 0.8.1 and up.

Answer #2:

I tried the proposed solution above:

df[df["A"].str.contains("Hello|Britain")]

and got an error:

ValueError: cannot mask with array containing NA / NaN values

you can transform NA values into False, like this:

df[df["A"].str.contains("Hello|Britain", na=False)]

Answer #3:

How do I select by partial string from a pandas DataFrame?

This post is meant for readers who want to

• search for a substring in a string column (the simplest case)
• search for multiple substrings (similar to isin)
• match a whole word from text (e.g., “blue” should match “the sky is blue” but not “bluejay”)
• match multiple whole words
• Understand the reason behind “ValueError: cannot index with vector containing NA / NaN values”

…and would like to know more about what methods should be preferred over others.

(P.S.: I’ve seen a lot of questions on similar topics, I thought it would be good to leave this here.)

Friendly disclaimer, this is post is long.

---

Basic Substring Search

# setup
df1 = pd.DataFrame({'col': ['foo', 'foobar', 'bar', 'baz']})
df1
      col
0     foo
1  foobar
2     bar
3     baz

str.contains can be used to perform either substring searches or regex based search. The search defaults to regex-based unless you explicitly disable it.

Here is an example of regex-based search,

# find rows in `df1` which contain "foo" followed by something
df1[df1['col'].str.contains(r'foo(?!$)')]
      col
1  foobar

Sometimes regex search is not required, so specify regex=False to disable it.

#select all rows containing "foo"
df1[df1['col'].str.contains('foo', regex=False)]
# same as df1[df1['col'].str.contains('foo')] but faster.
      col
0     foo
1  foobar

Performance wise, regex search is slower than substring search:

df2 = pd.concat([df1] * 1000, ignore_index=True)
%timeit df2[df2['col'].str.contains('foo')]
%timeit df2[df2['col'].str.contains('foo', regex=False)]
6.31 ms ± 126 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
2.8 ms ± 241 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Avoid using regex-based search if you don’t need it.

Addressing ValueErrors
Sometimes, performing a substring search and filtering on the result will result in

ValueError: cannot index with vector containing NA / NaN values

This is usually because of mixed data or NaNs in your object column,

s = pd.Series(['foo', 'foobar', np.nan, 'bar', 'baz', 123])
s.str.contains('foo|bar')
0     True
1     True
2      NaN
3     True
4    False
5      NaN
dtype: object
s[s.str.contains('foo|bar')]
# ---------------------------------------------------------------------------
# ValueError                                Traceback (most recent call last)

Anything that is not a string cannot have string methods applied on it, so the result is NaN (naturally). In this case, specify na=False to ignore non-string data,

s.str.contains('foo|bar', na=False)
0     True
1     True
2    False
3     True
4    False
5    False
dtype: bool

How do I apply this to multiple columns at once?
The answer is in the question. Use DataFrame.apply:

# `axis=1` tells `apply` to apply the lambda function column-wise.
df.apply(lambda col: col.str.contains('foo|bar', na=False), axis=1)
       A      B
0   True   True
1   True  False
2  False   True
3   True  False
4  False  False
5  False  False

All of the solutions below can be “applied” to multiple columns using the column-wise apply method (which is OK in my book, as long as you don’t have too many columns).

If you have a DataFrame with mixed columns and want to select only the object/string columns, take a look at select_dtypes.

---

Multiple Substring Search

This is most easily achieved through a regex search using the regex OR pipe.

# Slightly modified example.
df4 = pd.DataFrame({'col': ['foo abc', 'foobar xyz', 'bar32', 'baz 45']})
df4
          col
0     foo abc
1  foobar xyz
2       bar32
3      baz 45
df4[df4['col'].str.contains(r'foo|baz')]
          col
0     foo abc
1  foobar xyz
3      baz 45

You can also create a list of terms, then join them:

terms = ['foo', 'baz']
df4[df4['col'].str.contains('|'.join(terms))]
          col
0     foo abc
1  foobar xyz
3      baz 45

Sometimes, it is wise to escape your terms in case they have characters that can be interpreted as regex metacharacters. If your terms contain any of the following characters…

. ^ $ * + ? { } [ ]  | ( )

Then, you’ll need to use re.escape to escape them:

import re
df4[df4['col'].str.contains('|'.join(map(re.escape, terms)))]
          col
0     foo abc
1  foobar xyz
3      baz 45

re.escape has the effect of escaping the special characters so they’re treated literally.

re.escape(r'.foo^')
# '\.foo\^'

---

Matching Entire Word(s)

By default, the substring search searches for the specified substring/pattern regardless of whether it is full word or not. To only match full words, we will need to make use of regular expressions here—in particular, our pattern will need to specify word boundaries (b).

For example,

df3 = pd.DataFrame({'col': ['the sky is blue', 'bluejay by the window']})
df3
                     col
0        the sky is blue
1  bluejay by the window

Now consider,

df3[df3['col'].str.contains('blue')]
                     col
0        the sky is blue
1  bluejay by the window

v/s

df3[df3['col'].str.contains(r'bblueb')]
               col
0  the sky is blue

---

Multiple Whole Word Search

Similar to the above, except we add a word boundary (b) to the joined pattern.

p = r'b(?:{})b'.format('|'.join(map(re.escape, terms)))
df4[df4['col'].str.contains(p)]
       col
0  foo abc
3   baz 45

Where p looks like this,

p
# '\b(?:foo|baz)\b'

---

A Great Alternative: Use List Comprehensions!

Because you can! And you should! They are usually a little bit faster than string methods, because string methods are hard to vectorise and usually have loopy implementations.

Instead of,

df1[df1['col'].str.contains('foo', regex=False)]

Use the in operator inside a list comp,

df1[['foo' in x for x in df1['col']]]
       col
0  foo abc
1   foobar

Instead of,

regex_pattern = r'foo(?!$)'
df1[df1['col'].str.contains(regex_pattern)]

Use re.compile (to cache your regex) + Pattern.search inside a list comp,

p = re.compile(regex_pattern, flags=re.IGNORECASE)
df1[[bool(p.search(x)) for x in df1['col']]]
      col
1  foobar

If “col” has NaNs, then instead of

df1[df1['col'].str.contains(regex_pattern, na=False)]

Use,

def try_search(p, x):
    try:
        return bool(p.search(x))
    except TypeError:
        return False
p = re.compile(regex_pattern)
df1[[try_search(p, x) for x in df1['col']]]
      col
1  foobar

---

More Options for Partial String Matching: np.char.find, np.vectorize, DataFrame.query.

In addition to str.contains and list comprehensions, you can also use the following alternatives.

np.char.find
Supports substring searches (read: no regex) only.

df4[np.char.find(df4['col'].values.astype(str), 'foo') > -1]
          col
0     foo abc
1  foobar xyz

np.vectorize
This is a wrapper around a loop, but with lesser overhead than most pandas str methods.

f = np.vectorize(lambda haystack, needle: needle in haystack)
f(df1['col'], 'foo')
# array([ True,  True, False, False])
df1[f(df1['col'], 'foo')]
       col
0  foo abc
1   foobar

Regex solutions possible:

regex_pattern = r'foo(?!$)'
p = re.compile(regex_pattern)
f = np.vectorize(lambda x: pd.notna(x) and bool(p.search(x)))
df1[f(df1['col'])]
      col
1  foobar

DataFrame.query
Supports string methods through the python engine. This offers no visible performance benefits, but is nonetheless useful to know if you need to dynamically generate your queries.

df1.query('col.str.contains("foo")', engine='python')
      col
0     foo
1  foobar

More information on query and eval family of methods can be found at Dynamic Expression Evaluation in pandas using pd.eval().

---

Recommended Usage Precedence

1. (First) str.contains, for its simplicity and ease handling NaNs and mixed data
2. List comprehensions, for its performance (especially if your data is purely strings)
3. np.vectorize
4. (Last) df.query

Answer #4:

If anyone wonders how to perform a related problem: “Select column by partial string”

Use:

df.filter(like='hello')  # select columns which contain the word hello

And to select rows by partial string matching, pass axis=0 to filter:

# selects rows which contain the word hello in their index label
df.filter(like='hello', axis=0)

Answer #5:

Quick note: if you want to do selection based on a partial string contained in the index, try the following:

df['stridx']=df.index
df[df['stridx'].str.contains("Hello|Britain")]

Answer #6:

Say you have the following DataFrame:

>>> df = pd.DataFrame([['hello', 'hello world'], ['abcd', 'defg']], columns=['a','b'])
>>> df
       a            b
0  hello  hello world
1   abcd         defg

You can always use the in operator in a lambda expression to create your filter.

>>> df.apply(lambda x: x['a'] in x['b'], axis=1)
0     True
1    False
dtype: bool

The trick here is to use the axis=1 option in the apply to pass elements to the lambda function row by row, as opposed to column by column.

Answer #7:

Should you need to do a case insensitive search for a string in a pandas dataframe column:

df[df['A'].str.contains("hello", case=False)]

Answer #8:

Here’s what I ended up doing for partial string matches. If anyone has a more efficient way of doing this please let me know.

def stringSearchColumn_DataFrame(df, colName, regex):
    newdf = DataFrame()
    for idx, record in df[colName].iteritems():
        if re.search(regex, record):
            newdf = concat([df[df[colName] == record], newdf], ignore_index=True)
    return newdf