Related
I have a pandas DataFrame with a column of string values. I need to select rows based on partial string matches.
Something like this idiom:
re.search(pattern, cell_in_question)
returning a boolean. I am familiar with the syntax of df[df['A'] == "hello world"] but can't seem to find a way to do the same with a partial string match, say 'hello'.
Vectorized string methods (i.e. Series.str) let you do the following:
df[df['A'].str.contains("hello")]
This is available in pandas 0.8.1 and up.
I am using pandas 0.14.1 on macos in ipython notebook. I tried the proposed line above:
df[df["A"].str.contains("Hello|Britain")]
and got an error:
cannot index with vector containing NA / NaN values
but it worked perfectly when an "==True" condition was added, like this:
df[df['A'].str.contains("Hello|Britain")==True]
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) as in df1[df1['col'].str.contains(r'foo(?!$)')]
search for multiple substrings (similar to isin), e.g., with df4[df4['col'].str.contains(r'foo|baz')]
match a whole word from text (e.g., "blue" should match "the sky is blue" but not "bluejay"), e.g., with df3[df3['col'].str.contains(r'\bblue\b')]
match multiple whole words
Understand the reason behind "ValueError: cannot index with vector containing NA / NaN values" and correct it with str.contains('pattern',na=False)
...and would like to know more about what methods should be preferred over others.
(P.S.: I've seen a lot of questions on similar topics, I thought it would be good to leave this here.)
Friendly disclaimer, this is post is long.
Basic Substring Search
# setup
df1 = pd.DataFrame({'col': ['foo', 'foobar', 'bar', 'baz']})
df1
col
0 foo
1 foobar
2 bar
3 baz
str.contains can be used to perform either substring searches or regex based search. The search defaults to regex-based unless you explicitly disable it.
Here is an example of regex-based search,
# find rows in `df1` which contain "foo" followed by something
df1[df1['col'].str.contains(r'foo(?!$)')]
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'\bblue\b')]
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 Dynamically evaluate an expression from a formula in Pandas.
Recommended Usage Precedence
(First) str.contains, for its simplicity and ease handling NaNs and mixed data
List comprehensions, for its performance (especially if your data is purely strings)
np.vectorize
(Last) df.query
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)
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")]
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)]
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.
You can try considering them as string as :
df[df['A'].astype(str).str.contains("Hello|Britain")]
Suppose we have a column named "ENTITY" in the dataframe df. We can filter our df,to have the entire dataframe df, wherein rows of "entity" column doesn't contain "DM" by using a mask as follows:
mask = df['ENTITY'].str.contains('DM')
df = df.loc[~(mask)].copy(deep=True)
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
Using contains didn't work well for my string with special characters. Find worked though.
df[df['A'].str.find("hello") != -1]
A more generalised example - if looking for parts of a word OR specific words in a string:
df = pd.DataFrame([('cat andhat', 1000.0), ('hat', 2000000.0), ('the small dog', 1000.0), ('fog', 330000.0),('pet', 330000.0)], columns=['col1', 'col2'])
Specific parts of sentence or word:
searchfor = '.*cat.*hat.*|.*the.*dog.*'
Creat column showing the affected rows (can always filter out as necessary)
df["TrueFalse"]=df['col1'].str.contains(searchfor, regex=True)
col1 col2 TrueFalse
0 cat andhat 1000.0 True
1 hat 2000000.0 False
2 the small dog 1000.0 True
3 fog 330000.0 False
4 pet 3 30000.0 False
Maybe you want to search for some text in all columns of the Pandas dataframe, and not just in the subset of them. In this case, the following code will help.
df[df.apply(lambda row: row.astype(str).str.contains('String To Find').any(), axis=1)]
Warning. This method is relatively slow, albeit convenient.
Somewhat similar to #cs95's answer, but here you don't need to specify an engine:
df.query('A.str.contains("hello").values')
There are answers before this which accomplish the asked feature, anyway I would like to show the most generally way:
df.filter(regex=".*STRING_YOU_LOOK_FOR.*")
This way let's you get the column you look for whatever the way is wrote.
( Obviusly, you have to write the proper regex expression for each case )
My 2c worth:
I did the following:
sale_method = pd.DataFrame(model_data['Sale Method'].str.upper())
sale_method['sale_classification'] = \
np.where(sale_method['Sale Method'].isin(['PRIVATE']),
'private',
np.where(sale_method['Sale Method']
.str.contains('AUCTION'),
'auction',
'other'
)
)
df[df['A'].str.contains("hello", case=False)]
I have a pandas DataFrame with a column of string values. I need to select rows based on partial string matches.
Something like this idiom:
re.search(pattern, cell_in_question)
returning a boolean. I am familiar with the syntax of df[df['A'] == "hello world"] but can't seem to find a way to do the same with a partial string match, say 'hello'.
Vectorized string methods (i.e. Series.str) let you do the following:
df[df['A'].str.contains("hello")]
This is available in pandas 0.8.1 and up.
I am using pandas 0.14.1 on macos in ipython notebook. I tried the proposed line above:
df[df["A"].str.contains("Hello|Britain")]
and got an error:
cannot index with vector containing NA / NaN values
but it worked perfectly when an "==True" condition was added, like this:
df[df['A'].str.contains("Hello|Britain")==True]
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) as in df1[df1['col'].str.contains(r'foo(?!$)')]
search for multiple substrings (similar to isin), e.g., with df4[df4['col'].str.contains(r'foo|baz')]
match a whole word from text (e.g., "blue" should match "the sky is blue" but not "bluejay"), e.g., with df3[df3['col'].str.contains(r'\bblue\b')]
match multiple whole words
Understand the reason behind "ValueError: cannot index with vector containing NA / NaN values" and correct it with str.contains('pattern',na=False)
...and would like to know more about what methods should be preferred over others.
(P.S.: I've seen a lot of questions on similar topics, I thought it would be good to leave this here.)
Friendly disclaimer, this is post is long.
Basic Substring Search
# setup
df1 = pd.DataFrame({'col': ['foo', 'foobar', 'bar', 'baz']})
df1
col
0 foo
1 foobar
2 bar
3 baz
str.contains can be used to perform either substring searches or regex based search. The search defaults to regex-based unless you explicitly disable it.
Here is an example of regex-based search,
# find rows in `df1` which contain "foo" followed by something
df1[df1['col'].str.contains(r'foo(?!$)')]
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'\bblue\b')]
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 Dynamically evaluate an expression from a formula in Pandas.
Recommended Usage Precedence
(First) str.contains, for its simplicity and ease handling NaNs and mixed data
List comprehensions, for its performance (especially if your data is purely strings)
np.vectorize
(Last) df.query
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)
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")]
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)]
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.
You can try considering them as string as :
df[df['A'].astype(str).str.contains("Hello|Britain")]
Suppose we have a column named "ENTITY" in the dataframe df. We can filter our df,to have the entire dataframe df, wherein rows of "entity" column doesn't contain "DM" by using a mask as follows:
mask = df['ENTITY'].str.contains('DM')
df = df.loc[~(mask)].copy(deep=True)
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
Using contains didn't work well for my string with special characters. Find worked though.
df[df['A'].str.find("hello") != -1]
A more generalised example - if looking for parts of a word OR specific words in a string:
df = pd.DataFrame([('cat andhat', 1000.0), ('hat', 2000000.0), ('the small dog', 1000.0), ('fog', 330000.0),('pet', 330000.0)], columns=['col1', 'col2'])
Specific parts of sentence or word:
searchfor = '.*cat.*hat.*|.*the.*dog.*'
Creat column showing the affected rows (can always filter out as necessary)
df["TrueFalse"]=df['col1'].str.contains(searchfor, regex=True)
col1 col2 TrueFalse
0 cat andhat 1000.0 True
1 hat 2000000.0 False
2 the small dog 1000.0 True
3 fog 330000.0 False
4 pet 3 30000.0 False
Maybe you want to search for some text in all columns of the Pandas dataframe, and not just in the subset of them. In this case, the following code will help.
df[df.apply(lambda row: row.astype(str).str.contains('String To Find').any(), axis=1)]
Warning. This method is relatively slow, albeit convenient.
Somewhat similar to #cs95's answer, but here you don't need to specify an engine:
df.query('A.str.contains("hello").values')
There are answers before this which accomplish the asked feature, anyway I would like to show the most generally way:
df.filter(regex=".*STRING_YOU_LOOK_FOR.*")
This way let's you get the column you look for whatever the way is wrote.
( Obviusly, you have to write the proper regex expression for each case )
My 2c worth:
I did the following:
sale_method = pd.DataFrame(model_data['Sale Method'].str.upper())
sale_method['sale_classification'] = \
np.where(sale_method['Sale Method'].isin(['PRIVATE']),
'private',
np.where(sale_method['Sale Method']
.str.contains('AUCTION'),
'auction',
'other'
)
)
df[df['A'].str.contains("hello", case=False)]
Trying to make a Boolean flag that reads TRUE if one value or another is within a list. The below code is returning a FALSE for row 1 and I am not sure why, could someone help me understand why a FALSE is getting returned for the first row?
lists={'someList!':[[1,2,12,6,'ABC'],[1000,4,'z','a','bob']]}
dfLists = pd.DataFrame(lists)
dfLists['contains?']=dfLists['someList!'].isin([0,1])
Could someone help me understand why a FALSE is getting returned for the first row?
This isn't working because .isin(values) returns whether each element in the Series is contained in values.
You can use {0, 1} as a set and apply the truthiness of its intersection to each list:
>>> s = {0, 1}
>>> dfLists['someList!'].apply(lambda x: bool(s.intersection(x)))
0 True
1 False
This effectively does:
>>> s.intersection([1, 2, 12, 6, 'ABC'])
{1}
>>> s.intersection([1000, 4, 'z', 'a', 'bob'])
set()
The bool of the first result is True, because it is non-empty.
Using Dataframe constructor flatten you list column, then using isin
pd.DataFrame(dfLists['someList!'].tolist()).isin([1,2]).any(1)
Out[39]:
0 True
1 False
dtype: bool
You are passing the dataframe a list of lists. It's comparing integers to lists, so it's not finding a match.
Define your columns distinctly, like this, and isin should work.
lists={'someList!':[1,2,12,6,'ABC'], 'someList2':[1000,4,'z','a','bob']}
Since .ix has been deprecated as of Pandas 0.20, I wonder what is the proper way to mix lable-based, boolean-based and position-based indexing in Pandas? I need to assign values to a slice of dataframe that can be best referenced with label or boolean on the index and position on the columns. For example (using .loc as placeholder for the desired slicing method):
df.loc[df['a'] == 'x', -12:-1] = 3
obviously this doesn't work, with which I get:
TypeError: cannot do slice indexing on <class 'pandas.core.indexes.base.Index'> with these indexers [-12] of <class 'int'>
If I use .iloc, I get:
NotImplementedError: iLocation based boolean indexing on an integer type is not available
So how do I do it, without chaining, obviously to avoid chained assignment problem.
Let's use .loc with the boolean indexing, and accessing the column labels via the dataframe column index with index slicing:
df.loc[df['a'] == 'x', df.columns[-12:-1]] = 3
maybe I should've explained clearer. I meant if your dataframe is indexed (with 0 to n), then you can use loc[] for a combination of number for rows and lable for column:
new_df = pd.DataFrame({'a':[1,2,3,4],'b':[5,6,7,8]})
new_df
Out[10]:
a b
0 1 5
1 2 6
2 3 7
3 4 8
new_df.loc[0,'a']
Out[11]:
1
Even though .ix has been removed, it looks like .loc does the same job now. You can make a mix reference using .loc.
If you would like to filter those rows for which a string is in a column value, it is possible to use something like data.sample_id.str.contains('hph') (answered before: check if string in pandas dataframe column is in list, or Check if string is in a pandas dataframe).
However, my lookup column contains emtpy cells. Terefore, str.contains() yields NaN values and I get an value error upon indexing.
`ValueError: cannot index with vector containing NA / NaN values``
What works:
# get all runs
mask = [index for index, item in enumerate(data.sample_id.values) if 'zent' in str(item)]
Is there a more elegant and faster method (similar to str.contains()) than this one ?
You can set parameter na in str.contains to False:
print (df.a.str.contains('hph', na=False))
Using EdChum sample:
df = pd.DataFrame({'a':['hph', np.NaN, 'sadhphsad', 'hello']})
print (df)
a
0 hph
1 NaN
2 sadhphsad
3 hello
print (df.a.str.contains('hph', na=False))
0 True
1 False
2 True
3 False
Name: a, dtype: bool
IIUC you can filter those rows out also
data['sample'].dropna().str.contains('hph')
Example:
In [38]:
df =pd.DataFrame({'a':['hph', np.NaN, 'sadhphsad', 'hello']})
df
Out[38]:
a
0 hph
1 NaN
2 sadhphsad
3 hello
In [39]:
df['a'].dropna().str.contains('hph')
Out[39]:
0 True
2 True
3 False
Name: a, dtype: bool
So by calling dropna first you can then safely use str.contains on the Series as there will be no NaN values
Another way to handle the null values would be to use notnull:
In [43]:
(df['a'].notnull()) & (df['a'].str.contains('hph'))
Out[43]:
0 True
1 False
2 True
3 False
Name: a, dtype: bool
but I think passing na=False would be cleaner (#jezrael's answer)