I have this dataframe:
pd.DataFrame({'text': ['I have %%article%% car', '%%article%%fter dawn', 'D%%article%%t%%article%%Fr%%article%%me']})
I am trying to replace %%article%% by letter a to have as output:
pd.DataFrame({'text': ['I have a car', 'after dawn', 'DataFrame']})
I tried to create a dict ={'%%article%%':'a'} and then:
df['text'] = df['text'].map(dict)
But it's not working, it returns NaN
When passing a dict to Series.map, it uses table lookup so that only elements that exactly match '%%article%%' will be replaced by 'a'.
An example from doc:
>>> s = pd.Series(['cat', 'dog', np.nan, 'rabbit'])
>>> s
0 cat
1 dog
2 NaN
3 rabbit
>>> s.map({'cat': 'kitten', 'dog': 'puppy'})
0 kitten
1 puppy
2 NaN
3 NaN
An element with something like 'ccat' will not be replaced. Instead, you can use a function to replace them:
>>> df = pd.DataFrame({'text': ['I have %%article%% car', '%%article%%fter dawn', 'D%%article%%t%%article%%Fr%%article%%me']})
>>> df.text = df.text.map(lambda i: i.replace('%%article%%', 'a'))
>>> df
text
0 I have a car
1 after dawn
2 DataFrame
But the better is probably Series.replace:
>>> df.replace('%%article%%', 'a')
text
0 I have a car
1 after dawn
2 DataFrame
Use:
df['text'].str.replace('%%article%%', 'a')
Output:
0 I have a car
1 after dawn
2 DataFrame
Name: text, dtype: object
I have data saved in a postgreSQL database. I am querying this data using Python2.7 and turning it into a Pandas DataFrame. However, the last column of this dataframe has a dictionary of values inside it. The DataFrame df looks like this:
Station ID Pollutants
8809 {"a": "46", "b": "3", "c": "12"}
8810 {"a": "36", "b": "5", "c": "8"}
8811 {"b": "2", "c": "7"}
8812 {"c": "11"}
8813 {"a": "82", "c": "15"}
I need to split this column into separate columns, so that the DataFrame `df2 looks like this:
Station ID a b c
8809 46 3 12
8810 36 5 8
8811 NaN 2 7
8812 NaN NaN 11
8813 82 NaN 15
The major issue I'm having is that the lists are not the same lengths. But all of the lists only contain up to the same 3 values: 'a', 'b', and 'c'. And they always appear in the same order ('a' first, 'b' second, 'c' third).
The following code USED to work and return exactly what I wanted (df2).
objs = [df, pandas.DataFrame(df['Pollutant Levels'].tolist()).iloc[:, :3]]
df2 = pandas.concat(objs, axis=1).drop('Pollutant Levels', axis=1)
print(df2)
I was running this code just last week and it was working fine. But now my code is broken and I get this error from line [4]:
IndexError: out-of-bounds on slice (end)
I made no changes to the code but am now getting the error. I feel this is due to my method not being robust or proper.
Any suggestions or guidance on how to split this column of lists into separate columns would be super appreciated!
EDIT: I think the .tolist() and .apply methods are not working on my code because it is one Unicode string, i.e.:
#My data format
u{'a': '1', 'b': '2', 'c': '3'}
#and not
{u'a': '1', u'b': '2', u'c': '3'}
The data is imported from the postgreSQL database in this format. Any help or ideas with this issue? is there a way to convert the Unicode?
To convert the string to an actual dict, you can do df['Pollutant Levels'].map(eval). Afterwards, the solution below can be used to convert the dict to different columns.
Using a small example, you can use .apply(pd.Series):
In [2]: df = pd.DataFrame({'a':[1,2,3], 'b':[{'c':1}, {'d':3}, {'c':5, 'd':6}]})
In [3]: df
Out[3]:
a b
0 1 {u'c': 1}
1 2 {u'd': 3}
2 3 {u'c': 5, u'd': 6}
In [4]: df['b'].apply(pd.Series)
Out[4]:
c d
0 1.0 NaN
1 NaN 3.0
2 5.0 6.0
To combine it with the rest of the dataframe, you can concat the other columns with the above result:
In [7]: pd.concat([df.drop(['b'], axis=1), df['b'].apply(pd.Series)], axis=1)
Out[7]:
a c d
0 1 1.0 NaN
1 2 NaN 3.0
2 3 5.0 6.0
Using your code, this also works if I leave out the iloc part:
In [15]: pd.concat([df.drop('b', axis=1), pd.DataFrame(df['b'].tolist())], axis=1)
Out[15]:
a c d
0 1 1.0 NaN
1 2 NaN 3.0
2 3 5.0 6.0
I know the question is quite old, but I got here searching for answers. There is actually a better (and faster) way now of doing this using json_normalize:
import pandas as pd
df2 = pd.json_normalize(df['Pollutant Levels'])
This avoids costly apply functions...
The fastest method to normalize a column of flat, one-level dicts, as per the timing analysis performed by Shijith in this answer:
df.join(pd.DataFrame(df.pop('Pollutants').values.tolist()))
It will not resolve other issues, with columns of list or dicts, that are addressed below, such as rows with NaN, or nested dicts.
pd.json_normalize(df.Pollutants) is significantly faster than df.Pollutants.apply(pd.Series)
See the %%timeit below. For 1M rows, .json_normalize is 47 times faster than .apply.
Whether reading data from a file, or from an object returned by a database, or API, it may not be clear if the dict column has dict or str type.
If the dictionaries in the column are str type, they must be converted back to a dict type, using ast.literal_eval, or json.loads(…).
Use pd.json_normalize to convert the dicts, with keys as headers and values for rows.
There are additional parameters (e.g. record_path & meta) for dealing with nested dicts.
Use pandas.DataFrame.join to combine the original DataFrame, df, with the columns created using pd.json_normalize
If the index isn't integers (as in the example), first use df.reset_index() to get an index of integers, before doing the normalize and join.
pandas.DataFrame.pop is used to remove the specified column from the existing dataframe. This removes the need to drop the column later, using pandas.DataFrame.drop.
As a note, if the column has any NaN, they must be filled with an empty dict
df.Pollutants = df.Pollutants.fillna({i: {} for i in df.index})
If the 'Pollutants' column is strings, use '{}'.
Also see How to json_normalize a column with NaNs.
import pandas as pd
from ast import literal_eval
import numpy as np
data = {'Station ID': [8809, 8810, 8811, 8812, 8813, 8814],
'Pollutants': ['{"a": "46", "b": "3", "c": "12"}', '{"a": "36", "b": "5", "c": "8"}', '{"b": "2", "c": "7"}', '{"c": "11"}', '{"a": "82", "c": "15"}', np.nan]}
df = pd.DataFrame(data)
# display(df)
Station ID Pollutants
0 8809 {"a": "46", "b": "3", "c": "12"}
1 8810 {"a": "36", "b": "5", "c": "8"}
2 8811 {"b": "2", "c": "7"}
3 8812 {"c": "11"}
4 8813 {"a": "82", "c": "15"}
5 8814 NaN
# check the type of the first value in Pollutants
>>> print(type(df.iloc[0, 1]))
<class 'str'>
# replace NaN with '{}' if the column is strings, otherwise replace with {}
df.Pollutants = df.Pollutants.fillna('{}') # if the NaN is in a column of strings
# df.Pollutants = df.Pollutants.fillna({i: {} for i in df.index}) # if the column is not strings
# Convert the column of stringified dicts to dicts
# skip this line, if the column contains dicts
df.Pollutants = df.Pollutants.apply(literal_eval)
# reset the index if the index is not unique integers from 0 to n-1
# df.reset_index(inplace=True) # uncomment if needed
# remove and normalize the column of dictionaries, and join the result to df
df = df.join(pd.json_normalize(df.pop('Pollutants')))
# display(df)
Station ID a b c
0 8809 46 3 12
1 8810 36 5 8
2 8811 NaN 2 7
3 8812 NaN NaN 11
4 8813 82 NaN 15
5 8814 NaN NaN NaN
%%timeit
# dataframe with 1M rows
dfb = pd.concat([df]*20000).reset_index(drop=True)
%%timeit
dfb.join(pd.json_normalize(dfb.Pollutants))
[out]:
46.9 ms ± 201 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
%%timeit
pd.concat([dfb.drop(columns=['Pollutants']), dfb.Pollutants.apply(pd.Series)], axis=1)
[out]:
7.75 s ± 52.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
Try this: The data returned from SQL has to converted into a Dict.
or could it be "Pollutant Levels" is now Pollutants'
StationID Pollutants
0 8809 {"a":"46","b":"3","c":"12"}
1 8810 {"a":"36","b":"5","c":"8"}
2 8811 {"b":"2","c":"7"}
3 8812 {"c":"11"}
4 8813 {"a":"82","c":"15"}
df2["Pollutants"] = df2["Pollutants"].apply(lambda x : dict(eval(x)) )
df3 = df2["Pollutants"].apply(pd.Series )
a b c
0 46 3 12
1 36 5 8
2 NaN 2 7
3 NaN NaN 11
4 82 NaN 15
result = pd.concat([df, df3], axis=1).drop('Pollutants', axis=1)
result
StationID a b c
0 8809 46 3 12
1 8810 36 5 8
2 8811 NaN 2 7
3 8812 NaN NaN 11
4 8813 82 NaN 15
I strongly recommend the method extract the column 'Pollutants':
df_pollutants = pd.DataFrame(df['Pollutants'].values.tolist(), index=df.index)
it's much faster than
df_pollutants = df['Pollutants'].apply(pd.Series)
when the size of df is giant.
Merlin's answer is better and super easy, but we don't need a lambda function. The evaluation of dictionary can be safely ignored by either of the following two ways as illustrated below:
Way 1: Two steps
# step 1: convert the `Pollutants` column to Pandas dataframe series
df_pol_ps = data_df['Pollutants'].apply(pd.Series)
df_pol_ps:
a b c
0 46 3 12
1 36 5 8
2 NaN 2 7
3 NaN NaN 11
4 82 NaN 15
# step 2: concat columns `a, b, c` and drop/remove the `Pollutants`
df_final = pd.concat([df, df_pol_ps], axis = 1).drop('Pollutants', axis = 1)
df_final:
StationID a b c
0 8809 46 3 12
1 8810 36 5 8
2 8811 NaN 2 7
3 8812 NaN NaN 11
4 8813 82 NaN 15
Way 2: The above two steps can be combined in one go:
df_final = pd.concat([df, df['Pollutants'].apply(pd.Series)], axis = 1).drop('Pollutants', axis = 1)
df_final:
StationID a b c
0 8809 46 3 12
1 8810 36 5 8
2 8811 NaN 2 7
3 8812 NaN NaN 11
4 8813 82 NaN 15
Note : for dictionary with depth=1 (one-level)
>>> df
Station ID Pollutants
0 8809 {"a": "46", "b": "3", "c": "12"}
1 8810 {"a": "36", "b": "5", "c": "8"}
2 8811 {"b": "2", "c": "7"}
3 8812 {"c": "11"}
4 8813 {"a": "82", "c": "15"}
speed comparison for a large dataset of 10 million rows
>>> df = pd.concat([df]*2000000).reset_index(drop=True)
>>> print(df.shape)
(10000000, 2)
def apply_drop(df):
return df.join(df['Pollutants'].apply(pd.Series)).drop('Pollutants', axis=1)
def json_normalise_drop(df):
return df.join(pd.json_normalize(df.Pollutants)).drop('Pollutants', axis=1)
def tolist_drop(df):
return df.join(pd.DataFrame(df['Pollutants'].tolist())).drop('Pollutants', axis=1)
def vlues_tolist_drop(df):
return df.join(pd.DataFrame(df['Pollutants'].values.tolist())).drop('Pollutants', axis=1)
def pop_tolist(df):
return df.join(pd.DataFrame(df.pop('Pollutants').tolist()))
def pop_values_tolist(df):
return df.join(pd.DataFrame(df.pop('Pollutants').values.tolist()))
>>> %timeit apply_drop(df.copy())
1 loop, best of 3: 53min 20s per loop
>>> %timeit json_normalise_drop(df.copy())
1 loop, best of 3: 54.9 s per loop
>>> %timeit tolist_drop(df.copy())
1 loop, best of 3: 6.62 s per loop
>>> %timeit vlues_tolist_drop(df.copy())
1 loop, best of 3: 6.63 s per loop
>>> %timeit pop_tolist(df.copy())
1 loop, best of 3: 5.99 s per loop
>>> %timeit pop_values_tolist(df.copy())
1 loop, best of 3: 5.94 s per loop
+---------------------+-----------+
| apply_drop | 53min 20s |
| json_normalise_drop | 54.9 s |
| tolist_drop | 6.62 s |
| vlues_tolist_drop | 6.63 s |
| pop_tolist | 5.99 s |
| pop_values_tolist | 5.94 s |
+---------------------+-----------+
df.join(pd.DataFrame(df.pop('Pollutants').values.tolist())) is the fastest
How do I split a column of dictionaries into separate columns with pandas?
pd.DataFrame(df['val'].tolist()) is the canonical method for exploding a column of dictionaries
Here's your proof using a colorful graph.
Benchmarking code for reference.
Note that I am only timing the explosion since that's the most interesting part of answering this question - other aspects of result construction (such as whether to use pop or drop) are tangential to the discussion and can be ignored (it should be noted however that using pop avoids the followup drop call, so the final solution is a bit more performant, but we are still listifying the column and passing it to pd.DataFrame either way).
Additionally, pop destructively mutates the input DataFrame, making it harder to run in benchmarking code which assumes the input is not changed across test runs.
Critique of other solutions
df['val'].apply(pd.Series) is extremely slow for large N as pandas constructs Series objects for each row, then proceeds to construct a DataFrame from them. For larger N the performance dips to the order of minutes or hours.
pd.json_normalize(df['val'])) is slower simply because json_normalize is meant to work with a much more complex input data - particularly deeply nested JSON with multiple record paths and metadata. We have a simple flat dict for which pd.DataFrame suffices, so use that if your dicts are flat.
Some answers suggest df.pop('val').values.tolist() or df.pop('val').to_numpy().tolist(). I don't think it makes much of a difference whether you listify the series or the numpy array. It's one operation less to listify the series directly and really isn't slower so I'd recommend avoiding generating the numpy array in the intermediate step.
You can use join with pop + tolist. Performance is comparable to concat with drop + tolist, but some may find this syntax cleaner:
res = df.join(pd.DataFrame(df.pop('b').tolist()))
Benchmarking with other methods:
df = pd.DataFrame({'a':[1,2,3], 'b':[{'c':1}, {'d':3}, {'c':5, 'd':6}]})
def joris1(df):
return pd.concat([df.drop('b', axis=1), df['b'].apply(pd.Series)], axis=1)
def joris2(df):
return pd.concat([df.drop('b', axis=1), pd.DataFrame(df['b'].tolist())], axis=1)
def jpp(df):
return df.join(pd.DataFrame(df.pop('b').tolist()))
df = pd.concat([df]*1000, ignore_index=True)
%timeit joris1(df.copy()) # 1.33 s per loop
%timeit joris2(df.copy()) # 7.42 ms per loop
%timeit jpp(df.copy()) # 7.68 ms per loop
One line solution is following:
>>> df = pd.concat([df['Station ID'], df['Pollutants'].apply(pd.Series)], axis=1)
>>> print(df)
Station ID a b c
0 8809 46 3 12
1 8810 36 5 8
2 8811 NaN 2 7
3 8812 NaN NaN 11
4 8813 82 NaN 15
df = pd.concat([df['a'], df.b.apply(pd.Series)], axis=1)
I've concatenated those steps in a method, you have to pass only the dataframe and the column which contains the dict to expand:
def expand_dataframe(dw: pd.DataFrame, column_to_expand: str) -> pd.DataFrame:
"""
dw: DataFrame with some column which contain a dict to expand
in columns
column_to_expand: String with column name of dw
"""
import pandas as pd
def convert_to_dict(sequence: str) -> Dict:
import json
s = sequence
json_acceptable_string = s.replace("'", "\"")
d = json.loads(json_acceptable_string)
return d
expanded_dataframe = pd.concat([dw.drop([column_to_expand], axis=1),
dw[column_to_expand]
.apply(convert_to_dict)
.apply(pd.Series)],
axis=1)
return expanded_dataframe
my_df = pd.DataFrame.from_dict(my_dict, orient='index', columns=['my_col'])
.. would have parsed the dict properly (putting each dict key into a separate df column, and key values into df rows), so the dicts would not get squashed into a single column in the first place.
I have a column called users in dataframe which doesn't have a unique format. I am doing a data cleanup project as the data looks unreadable.
company Users
A [{"Name":"Martin","Email":"name_1#email.com","EmpType":"Full"},{"Name":"Rick","Email":"name_2#email.com","Dept":"HR"}]
B [{"Name":"John","Email":"name_2#email.com","EmpType":"Full","Dept":"Sales" }]
I used the below query to this has broke down the data frame as below
df2 = df
df2 = df2.join(df['Users_config'].str.split('},{', expand=True).add_prefix('Users'))
company Users0 Users1
A "Name":"Martin","Email":"name_1#email.com","EmpType":"Full" "Name":"Rick","Email":"name_2#email.com","Dept":"HR"
B "Name":"John","Email":"name_2#email.com","EmpType":"Full","Dept":"Sales"
and further breaking the above df with "," using the same query I got the output as
Company Users01 Users02 Users03 Users10 Users11 Users12
1 "Name":"Martin" "Email":"name_1#email.com" "EmpType":"Full" "Name":"Rick" "Email":"name_2#email.com" "Dept":"HR"
2 "Name":"John" "Email":"name_2#email.com" "EmpType":"Full" "Dept":"Sales"
As this dataframe looks messy I want to get the output as below. I feel the best way to name the column is to use the column value "Name" from "Name":"Martin" itself and If we hardcore using df.rename the column name will get mismatch.
Company Name_1 Email_1 EmpType_1 Dept_1 Name_2 Email_2 Dept_2
1 Martin name_1#email.com Full Rick name_2#email.com "HR"
2 John name_2#email.com" Full Sales
Is there any way I can get the above output from the original dataframe.
Use:
df['Users'] = df['Users'].apply(ast.literal_eval)
d = df.explode('Users').reset_index(drop=True)
d = d.join(pd.DataFrame(d.pop('Users').tolist()))
d = d.set_index(['company', d.groupby('company').cumcount().add(1).astype(str)]).unstack()
d.columns = d.columns.map('_'.join)
Details:
First we use ast.literal_eval to evaluate the strings in Users column, then use DataFrame.explode on column Users to create a dataframe d.
print(d)
company Users
0 A {'Name': 'Martin', 'Email': 'name_1#email.com', 'EmpType': 'Full'}
1 A {'Name': 'Rick', 'Email': 'name_2#email.com', 'Dept': 'HR'}
2 B {'Name': 'John', 'Email': 'name_2#email.com', 'EmpType': 'Full', 'Dept': 'Sales'}
Create a new dataframe from the Users column in d and use DataFrame.join to join this new dataframe with d.
print(d)
company Name Email EmpType Dept
0 A Martin name_1#email.com Full NaN
1 A Rick name_2#email.com NaN HR
2 B John name_2#email.com Full Sales
Use DataFrame.groupby on column company then use groupby.cumcount to create a counter for each group, then use DataFrame.set_index to set the index of d as company + counter. Then use DataFrame.unstack to reshape the dataframe creating MultiIndex columns.
print(d)
Name Email EmpType Dept
1 2 1 2 1 2 1 2
company
A Martin Rick name_1#email.com name_2#email.com Full NaN NaN HR
B John NaN name_2#email.com NaN Full NaN Sales NaN
Finally use map along with .join to flatten the MultiIndex columns.
print(d)
Name_1 Name_2 Email_1 Email_2 EmpType_1 EmpType_2 Dept_1 Dept_2
company
A Martin Rick name_1#email.com name_2#email.com Full NaN NaN HR
B John NaN name_2#email.com NaN Full NaN Sales NaN
I have a Pandas DataFrame with two columns of "complementary" data. For any given row, there are 3 possibilities:
1) Column A has a non-null value, and column B has a null value, NaN, that I want to replace with the non-null value from column A.
2) Column A has a null value, NaN, that I want to replace with the non-null value from column B.
3) Both columns A and B have null values, NaN, which means I'll keep NaN as the value for that row.
Here's a simplified version of my DataFrame:
df1 = pd.DataFrame({'A' : ['keep1', np.nan, np.nan, 'keep4', np.nan],
'B' : [np.nan, 'keep2', np.nan, np.nan, np.nan]})
I was thinking that as an intermediate step, I'd create a new column C with the entries I need:
df2 = pd.DataFrame({'A' : ['keep1', np.nan, np.nan, 'keep4', np.nan],
'B' : [np.nan, 'keep2', np.nan, np.nan, np.nan],
'C' : ['keep1', 'keep2', np.nan, 'keep4', np.nan]}
Then I'd drop the first two rows A and B:
df_final = df2.drop(['A', 'B'], axis=1)
My actual DataFrame has hundreds of rows, and I've tried several approaches (boolean filters, looping through the DataFrame using iterrows, using DataFrame.where()) without success. I'd think this would be a simple problem, but I'm not seeing it. Any help is appreciated.
Thanks
You can use combine_first() to fill the gaps in A from B:
df1['C'] = df1['A'].combine_first(df1['B'])
#0 keep1
#1 keep2
#2 NaN
#3 keep4
#4 NaN
Use Series.fillna for replace missing values from A by B values:
df1['C'] = df1.A.fillna(df1.B)
print (df1)
A B C
0 keep1 NaN keep1
1 NaN keep2 keep2
2 NaN NaN NaN
3 keep4 NaN keep4
4 NaN NaN NaN
For avoid drop is possible use DataFrame.pop for extract columns:
df1['C'] = df1.pop('A').fillna(df1.pop('B'))
print (df1)
C
0 keep1
1 keep2
2 NaN
3 keep4
4 NaN