As Title says, I'm looking for a perfect solution to replace exact string in a series ignoring case.
ls = {'CAT':'abc','DOG' : 'def','POT':'ety'}
d = pd.DataFrame({'Data': ['cat','dog','pot','Truncate','HotDog','ShuPot'],'Result':['abc','def','ety','Truncate','HotDog','ShuPot']})
d
In the above code, ref hold the key-value pair where key is the existing value in a dataframe column and value is value to replace with.
Issue with this case is, service that pass the dictionary always holds dictionary key in upper case where dataframe might have value in lowercase.
expected output is stored in 'Result Column.
I tried including re.ignore = True which changes the last 2 values.
following code but that is not working as expected. it also converting values to upper case from previous iteration.
for k,v in ls.items():
print (k,v)
d['Data'] = d['Data'].astype(str).str.upper().replace({k:v})
print (d)
I'd appreciate any help.
Create a mapping series from the given dictionary, then transform the index of the mapping series to lower case, then using Series.map map the values in Data column to the values in mappings, then use Series.fillna to fill the missing values in the mapped series:
mappings = pd.Series(ls)
mappings.index = mappings.index.str.lower()
d['Result'] = d['Data'].str.lower().map(mappings).fillna(d['Data'])
# print(d)
Data Result
0 cat abc
1 dog def
2 pot ety
3 Truncate Truncate
4 HotDog HotDog
5 ShuPot ShuPot
Related
I am extracting tables from pdf using Camelot. Two of the columns are getting merged together with a newline separator. Is there a way to separate them into two columns?
Suppose the column looks like this.
A\nB
1\n2
2\n3
3\n4
Desired output:
|A|B|
|-|-|
|1|2|
|2|3|
|3|4|
I have tried df['A\nB'].str.split('\n', 2, expand=True) and that splits it into two columns however I want the new column names to be A and B and not 0 and 1. Also I need to pass a generalized column label instead of actual column name since I need to implement this for several docs which may have different column names. I can determine such column name in my dataframe using
colNew = df.columns[df.columns.str.contains(pat = '\n')]
However when I pass colNew in split function, it throws an attribute error
df[colNew].str.split('\n', 2, expand=True)
AttributeError: DataFrame object has no attribute 'str'
You can take advantage of the Pandas split function.
import pandas as pd
# recreate your pandas series above.
df = pd.DataFrame({'A\nB':['1\n2','2\n3','3\n4']})
# first: Turn the col into str.
# second. split the col based on seperator \n
# third: make sure expand as True since you want the after split col become two new col
test = df['A\nB'].astype('str').str.split('\n',expand=True)
# some rename
test.columns = ['A','B']
I hope this is helpful.
I reproduced the error from my side... I guess the issue is that "df[colNew]" is still a dataframe as it contains the indexes.
But .str.split() only works on Series. So taking as example your code, I would convert the dataframe to series using iloc[:,0].
Then another line to split the column headers:
df2=df[colNew].iloc[:,0].str.split('\n', 2, expand=True)
df2.columns = 'A\nB'.split('\n')
I have a column in a dataframe where all the values are lists (list of one item usually for each row). So, I would like to use get_dummies to one hot encode all the values. However, there may be a few rows where there is not a value for the column. I have seen it originally as a nan and then I have replaced that nan with an empty list, but in either case I do not see 0 and 1s for the result for the get_dummies, but rather each generated column is blank (I would expect each generated column to be 0).
How do I get get_dummies to work with an empty list?
# create column from dict where value will be a list
X['sponsor_list'] = X['bill_id'].map(sponsor_non_plaw_dict)
# line to replace nan in sponsor_list column with empty list
X.loc[X['sponsor_list'].isnull(),['sponsor_list']] = X.loc[X['sponsor_list'].isnull(),'sponsor_list'].apply(lambda x: [])
# use of get_dummies to encode the sponsor_list column
X = pd.concat([X, pd.get_dummies(X.sponsor_list.apply(pd.Series).stack()).sum(level=0)], axis=1)
Example:
111th-congress_senate-bill_3695.txt False ['Menendez,_Robert_[D-NJ].txt']
112th-congress_house-bill_3630.txt False []
111th-congress_senate-bill_852.txt False ['Vitter,_David_[R-LA].txt']
114th-congress_senate-bill_2832.txt False
['Isakson,_Johnny_[R-GA].txt']
107th-congress_senate-bill_535.txt False ['Bingaman,_Jeff_[D-NM].txt']
I want to one hot encode on the third column. That particular data item in the 2nd row has no person associated it with them, so I need that row to be encoded with all 0s. The reason I need the third column to be a list is because I need to do this to a related column as well where I need to have [0,n] values where n can be 5 or 10 or even 20.
X['sponsor_list'] = X['bill_id'].map(sponsor_non_plaw_dict)
X.loc[X['sponsor_list'].isnull(),['sponsor_list']] = X.loc[X['sponsor_list'].isnull(),'sponsor_list'].apply(lambda x: [])
mlb = MultiLabelBinarizer()
X = X.join(pd.DataFrame(mlb.fit_transform(X.pop('sponsor_list')),
columns=mlb.classes_,
index=X.index))
I used a MultiLabelBinarizer to capture what I was trying to do. I still replace nan with empty list before applying, but then I fit_transform to create the 0/1 values which can result in no 1's in a row, or many 1's in a row.
I have a very large data frame that I want to split ALL of the columns except first two based on a comma delimiter. So I need to logically reference column names in a loop or some other way to split all the columns in one swoop.
In my testing of the split method:
I have been able to explicitly refer to ( i.e. HARD CODE) a single column name (rs145629793) as one of the required parameters and the result was 2 new columns as I wanted.
See python code below
HARDCODED COLUMN NAME --
df[['rs1','rs2']] = df.rs145629793.str.split(",", expand = True)
The problem:
It is not feasible to refer to the actual column names and repeat code.
I then replaced the actual column name rs145629793 with columns[2] in the split method parameter list.
It results in an ERROR
'str has ni str attribute'
You can index columns by position rather than name using iloc. For example, to get the third column:
df.iloc[:, 2]
Thus you can easily loop over the columns you need.
I know what you are asking, but it's still helpful to provide some input data and expected output data. I have included random input data in my code below, so you can just copy and paste this to run, and try to apply it to your dataframe:
import pandas as pd
your_dataframe=pd.DataFrame({'a':['1,2,3', '9,8,7'],
'b':['4,5,6', '6,5,4'],
'c':['7,8,9', '3,2,1']})
import copy
def split_cols(df):
dict_of_df = {}
cols=df.columns.to_list()
for col in cols:
key_name = 'df'+str(col)
dict_of_df[key_name] = copy.deepcopy(df)
var=df[col].str.split(',', expand=True).add_prefix(col)
df=pd.merge(df, var, how='left', left_index=True, right_index=True).drop(col, axis=1)
return df
split_cols(your_dataframe)
Essentially, in this solution you create a list of the columns that you want to loop through. Then you loop through that list and create new dataframes for each column where you run the split() function. Then you merge everything back together on the index. I also:
included a prefix of the column name, so the column names did not have duplicate names and could be more easily identifiable
dropped the old column that we did the split on.
Just import copy and use the split_cols() function that I have created and pass the name of your dataframe.
I'm trying to replace the values in one column of a dataframe. The column ('female') only contains the values 'female' and 'male'.
I have tried the following:
w['female']['female']='1'
w['female']['male']='0'
But receive the exact same copy of the previous results.
I would ideally like to get some output which resembles the following loop element-wise.
if w['female'] =='female':
w['female'] = '1';
else:
w['female'] = '0';
I've looked through the gotchas documentation (http://pandas.pydata.org/pandas-docs/stable/gotchas.html) but cannot figure out why nothing happens.
Any help will be appreciated.
If I understand right, you want something like this:
w['female'] = w['female'].map({'female': 1, 'male': 0})
(Here I convert the values to numbers instead of strings containing numbers. You can convert them to "1" and "0", if you really want, but I'm not sure why you'd want that.)
The reason your code doesn't work is because using ['female'] on a column (the second 'female' in your w['female']['female']) doesn't mean "select rows where the value is 'female'". It means to select rows where the index is 'female', of which there may not be any in your DataFrame.
You can edit a subset of a dataframe by using loc:
df.loc[<row selection>, <column selection>]
In this case:
w.loc[w.female != 'female', 'female'] = 0
w.loc[w.female == 'female', 'female'] = 1
w.female.replace(to_replace=dict(female=1, male=0), inplace=True)
See pandas.DataFrame.replace() docs.
Slight variation:
w.female.replace(['male', 'female'], [1, 0], inplace=True)
This should also work:
w.female[w.female == 'female'] = 1
w.female[w.female == 'male'] = 0
This is very compact:
w['female'][w['female'] == 'female']=1
w['female'][w['female'] == 'male']=0
Another good one:
w['female'] = w['female'].replace(regex='female', value=1)
w['female'] = w['female'].replace(regex='male', value=0)
You can also use apply with .get i.e.
w['female'] = w['female'].apply({'male':0, 'female':1}.get):
w = pd.DataFrame({'female':['female','male','female']})
print(w)
Dataframe w:
female
0 female
1 male
2 female
Using apply to replace values from the dictionary:
w['female'] = w['female'].apply({'male':0, 'female':1}.get)
print(w)
Result:
female
0 1
1 0
2 1
Note: apply with dictionary should be used if all the possible values of the columns in the dataframe are defined in the dictionary else, it will have empty for those not defined in dictionary.
Using Series.map with Series.fillna
If your column contains more strings than only female and male, Series.map will fail in this case since it will return NaN for other values.
That's why we have to chain it with fillna:
Example why .map fails:
df = pd.DataFrame({'female':['male', 'female', 'female', 'male', 'other', 'other']})
female
0 male
1 female
2 female
3 male
4 other
5 other
df['female'].map({'female': '1', 'male': '0'})
0 0
1 1
2 1
3 0
4 NaN
5 NaN
Name: female, dtype: object
For the correct method, we chain map with fillna, so we fill the NaN with values from the original column:
df['female'].map({'female': '1', 'male': '0'}).fillna(df['female'])
0 0
1 1
2 1
3 0
4 other
5 other
Name: female, dtype: object
Alternatively there is the built-in function pd.get_dummies for these kinds of assignments:
w['female'] = pd.get_dummies(w['female'],drop_first = True)
This gives you a data frame with two columns, one for each value that occurs in w['female'], of which you drop the first (because you can infer it from the one that is left). The new column is automatically named as the string that you replaced.
This is especially useful if you have categorical variables with more than two possible values. This function creates as many dummy variables needed to distinguish between all cases. Be careful then that you don't assign the entire data frame to a single column, but instead, if w['female'] could be 'male', 'female' or 'neutral', do something like this:
w = pd.concat([w, pd.get_dummies(w['female'], drop_first = True)], axis = 1])
w.drop('female', axis = 1, inplace = True)
Then you are left with two new columns giving you the dummy coding of 'female' and you got rid of the column with the strings.
w.replace({'female':{'female':1, 'male':0}}, inplace = True)
The above code will replace 'female' with 1 and 'male' with 0, only in the column 'female'
There is also a function in pandas called factorize which you can use to automatically do this type of work. It converts labels to numbers: ['male', 'female', 'male'] -> [0, 1, 0]. See this answer for more information.
w.female = np.where(w.female=='female', 1, 0)
if someone is looking for a numpy solution. This is useful to replace values based on a condition. Both if and else conditions are inherent in np.where(). The solutions that use df.replace() may not be feasible if the column included many unique values in addition to 'male', all of which should be replaced with 0.
Another solution is to use df.where() and df.mask() in succession. This is because neither of them implements an else condition.
w.female.where(w.female=='female', 0, inplace=True) # replace where condition is False
w.female.mask(w.female=='female', 1, inplace=True) # replace where condition is True
dic = {'female':1, 'male':0}
w['female'] = w['female'].replace(dic)
.replace has as argument a dictionary in which you may change and do whatever you want or need.
I think that in answer should be pointed which type of object do you get in all methods suggested above: is it Series or DataFrame.
When you get column by w.female. or w[[2]] (where, suppose, 2 is number of your column) you'll get back DataFrame.
So in this case you can use DataFrame methods like .replace.
When you use .loc or iloc you get back Series, and Series don't have .replace method, so you should use methods like apply, map and so on.
To answer the question more generically so it applies to more use cases than just what the OP asked, consider this solution. I used jfs's solution solution to help me. Here, we create two functions that help feed each other and can be used whether you know the exact replacements or not.
import numpy as np
import pandas as pd
class Utility:
#staticmethod
def rename_values_in_column(column: pd.Series, name_changes: dict = None) -> pd.Series:
"""
Renames the distinct names in a column. If no dictionary is provided for the exact name changes, it will default
to <column_name>_count. Ex. female_1, female_2, etc.
:param column: The column in your dataframe you would like to alter.
:param name_changes: A dictionary of the old values to the new values you would like to change.
Ex. {1234: "User A"} This would change all occurrences of 1234 to the string "User A" and leave the other values as they were.
By default, this is an empty dictionary.
:return: The same column with the replaced values
"""
name_changes = name_changes if name_changes else {}
new_column = column.replace(to_replace=name_changes)
return new_column
#staticmethod
def create_unique_values_for_column(column: pd.Series, except_values: list = None) -> dict:
"""
Creates a dictionary where the key is the existing column item and the value is the new item to replace it.
The returned dictionary can then be passed the pandas rename function to rename all the distinct values in a
column.
Ex. column ["statement"]["I", "am", "old"] would return
{"I": "statement_1", "am": "statement_2", "old": "statement_3"}
If you would like a value to remain the same, enter the values you would like to stay in the except_values.
Ex. except_values = ["I", "am"]
column ["statement"]["I", "am", "old"] would return
{"old", "statement_3"}
:param column: A pandas Series for the column with the values to replace.
:param except_values: A list of values you do not want to have changed.
:return: A dictionary that maps the old values their respective new values.
"""
except_values = except_values if except_values else []
column_name = column.name
distinct_values = np.unique(column)
name_mappings = {}
count = 1
for value in distinct_values:
if value not in except_values:
name_mappings[value] = f"{column_name}_{count}"
count += 1
return name_mappings
For the OP's use case, it is simple enough to just use
w["female"] = Utility.rename_values_in_column(w["female"], name_changes = {"female": 0, "male":1}
However, it is not always so easy to know all of the different unique values within a data frame that you may want to rename. In my case, the string values for a column are hashed values so they hurt the readability. What I do instead is replace those hashed values with more readable strings thanks to the create_unique_values_for_column function.
df["user"] = Utility.rename_values_in_column(
df["user"],
Utility.create_unique_values_for_column(df["user"])
)
This will changed my user column values from ["1a2b3c", "a12b3c","1a2b3c"] to ["user_1", "user_2", "user_1]. Much easier to compare, right?
If you have only two classes you can use equality operator. For example:
df = pd.DataFrame({'col1':['a', 'a', 'a', 'b']})
df['col1'].eq('a').astype(int)
# (df['col1'] == 'a').astype(int)
Output:
0 1
1 1
2 1
3 0
Name: col1, dtype: int64
n1 = DataFrame({'zhanghui':[1,2,3,4] , 'wudi':[17,'gx',356,23] ,'sas'[234,51,354,123] })
n2 = DataFrame({'zhanghui_x':[1,2,3,5] , 'wudi':[17,23,'sd',23] ,'wudi_x':[17,23,'x356',23] ,'wudi_y':[17,23,'y356',23] ,'ddd':[234,51,354,123] })
code above defined two DataFrame objects. I wanna use 'zhanghui' field from n1 and 'zhanghui_x' field from n2 as "on" field merge n1 and n2,so my code like this:
n1.merge(n2,how = 'inner',left_on = 'zhanghui',right_on='zhanghui_x')
and then result columns given like this :
sas wudi_x zhanghui ddd wudi_y wudi_x wudi_y zhanghui_x
Some duplicate columns appeared,such as 'wudi_x' ,'wudi_y'.
So it's a pandas inner problems or I had a wrong usage about pd.merge ?
From pandas documentation, the merge() function has following properties;
pd.merge(left, right, how='inner', on=None, left_on=None, right_on=None,
left_index=False, right_index=False, sort=True,
suffixes=('_x', '_y'), copy=True, indicator=False,
validate=None)
where suffixes denote default suffix string to be attached to 'over-lapping' columns with defaults '_x' and '_y'.
I'm not sure if I understood your follow-up question correctly, but;
#case1
if the first dataFrame has column 'column_name_x' and the second dataFrame has column 'column_name' then there are no over-lapping columns and therefore no suffixes are attached.
#case2
if the first dataFrame has columns 'column_name', 'column_name_x' and the second dataFrame also has column 'column_name', the default suffixes attach to over-lapping columns and therefore the first frame's 'columnn_name' becomes 'column_name_x' and result in a duplicate of already existing column.
You can however, pass a None value to one(not all) of the suffixes to ensure that column names of certain dataFrame remain as-is.
Your approach is right, pandas automatically gives postscripts after merging the columns that are "duplicated" with the original headers given a postscript _x, _y, etc.
you can first select what columns to merge and proceed:
cols_to_use = n2.columns - n1.columns
n1.merge(n2[cols_to_use],how = 'inner',left_on = 'zhanghui',right_on='zhanghui_x')
result columns:
sas wudi zhanghui ddd wudi_x wudi_y zhanghui_x
When I tried to run cols_to_use = n2.columns - n1.columns,it gave me a TypeError like this:
cannot perform __sub__ with this index type: <class pandas.core.indexes.base.Index'>
then I tried to use code below:
cols_to_use = [i for i in list(n2.columns) if i not in list(n1.columns) ]
It worked fine,result columns given like this:
sas wudi zhanghui ddd wudi_x wudi_y zhanghui_x
So,#S Ringne's method really resolved my problems.
=============================================
Pandas just simply add suffix such as '_x' to resolve the duplicate-column-name problem when it comes to merging two Frame objects.
But what will it happen if the name form of 'a-column-name'+'_x' appears in either Frame object? I used to think that it will check if the name form of 'a-column-name'+'_x' appears, But actually pandas doesn't have this check?