I have a dataframe that has 20 or so columns in it. One of the columns is called 'director_name' and has values such as 'John Doe' or 'Jane Doe'. I want to split this into 2 columns, 'First_Name' and 'Last_Name'. When I run the following it works as expected and splits the string into 2 columns:
data[['First_Name', 'Last_Name']] = data.director_name.str.split(' ', expand
= True)
data
First_Name Last_Name
John Doe
It works great, however it does NOT work when I have NULL (NaN) values under 'director_name'. It throws the following error:
'Columns must be same length as key'
I'd like to add a function which checks if the value != null, then do the command listed above, otherwise enter 'NA' for First_Name and 'Last_Name'
Any ideas how I would go about that?
EDIT:
I just checked the file and I'm not sure if NULL is the issue. I have some names that are 3-4 strings long. i.e.
John Allen Doe
John Allen Doe Jr
Maybe I can't split this into First_Name and Last_Name.
Hmmmm
Here is a way is to split and choose say the first two values as first name and last name
Id name
0 1 James Cameron
1 2 Martin Sheen
2 3 John Allen Doe
3 4 NaN
df['First_Name'] = df.name.str.split(' ', expand = True)[0]
df['Last_Name'] = df.name.str.split(' ', expand = True)[1]
You get
Id name First_Name Last_Name
0 1 James Cameron James Cameron
1 2 Martin Sheen Martin Sheen
2 3 John Allen Doe John Allen
3 4 NaN NaN None
Use str.split (no parameter, because splitter by default whitespace) with indexing with str for select lists by position:
print (df.name.str.split())
0 [James, Cameron]
1 [Martin, Sheen]
2 [John, Allen, Doe]
3 NaN
Name: name, dtype: object
df['First_Name'] = df.name.str.split().str[0]
df['Last_Name'] = df.name.str.split().str[1]
#data borrow from A-Za-z answer
print (df)
Id name First_Name Last_Name
0 1 James Cameron James Cameron
1 2 Martin Sheen Martin Sheen
2 3 John Allen Doe John Allen
3 4 NaN NaN NaN
There is also possible use paramter n for selecting second or first 2 names:
df['First_Name'] = df.name.str.split().str[0]
df['Last_Name'] = df.name.str.split(n=1).str[1]
print (df)
Id name First_Name Last_Name
0 1 James Cameron James Cameron
1 2 Martin Sheen Martin Sheen
2 3 John Allen Doe John Allen Doe
3 4 NaN NaN NaN
Solution with str.rstrip
df['First_Name'] = df.name.str.rsplit(n=1).str[0]
df['Last_Name'] = df.name.str.rsplit().str[-1]
print (df)
Id name First_Name Last_Name
0 1 James Cameron James Cameron
1 2 Martin Sheen Martin Sheen
2 3 John Allen Doe John Allen Doe
3 4 NaN NaN NaN
df['First_Name'] = df.name.str.split(' ', expand = True)[0]
df['Last_Name'] = df.name.str.split(' ', expand = True)[1]
This should do
This should fix your problem
Setup
data= pd.DataFrame({'director_name': {0: 'John Doe', 1: np.nan, 2: 'Alan Smith'}})
data
Out[457]:
director_name
0 John Doe
1 NaN
2 Alan Smith
Solution
#use a lambda function to check nan before splitting the column.
data[['First_Name', 'Last_Name']] = data.apply(lambda x: pd.Series([np.nan,np.nan] if pd.isnull(x.director_name) else x.director_name.split()), axis=1)
data
Out[446]:
director_name First_Name Last_Name
0 John Doe John Doe
1 NaN NaN NaN
2 Alan Smith Alan Smith
If you need to take only the first 2 names, you can do:
data[['First_Name', 'Last_Name']] = data.apply(lambda x: pd.Series([np.nan,np.nan] if pd.isnull(x.director_name) else x.director_name.split()).iloc[:2], axis=1)
Related
I am trying to create a relationship between two data frames that are related, but there is no key that creates a relationship. Here is the layout of my problem:
The first data frame that I am using is information about when people entered an amusement park. In this amusement park, people can stay at the park for multiple days. So the structure of this data frame is
id
name
date
0
John Smith
07-01-2020 10:13:24
1
John Smith
07-22-2020 09:47:04
4
Jane Doe
07-22-2020 09:47:04
2
Jane Doe
06-13-2020 13:27:53
3
Thomas Wallace
07-08-2020 11:15:28
So people may visit the park once, or multiple times (assume that name is a unique identifier for people). For the other data frame, the data is what rides they went on during their time at the park. So the structure of this data frame is
name
ride
date
John Smith
Insanity
07-01-2020 13:53:07
John Smith
Bumper Cars
07-01-2020 16:37:29
John Smith
Tilt-A-Whirl
07-02-2020 08:21:18
John Smith
Insanity
07-22-2020 11:44:32
Jane Doe
Bumper Cars
06-13-2020 14:14:41
Jane Doe
Teacups
06-13-2020 17:31:56
Thomas Wallace
Insanity
07-08-2020 13:20:23
With these two data frames, I want to get the id of the visit associated with the rides that they went on during that visit. So the desired output in this example would be
id
name
ride
date
0
John Smith
Insanity
07-01-2020 13:53:07
0
John Smith
Bumper Cars
07-01-2020 16:37:29
0
John Smith
Tilt-A-Whirl
07-02-2020 08:21:18
1
John Smith
Insanity
07-22-2020 11:44:32
2
Jane Doe
Bumper Cars
06-13-2020 14:14:41
2
Jane Doe
Teacups
06-13-2020 17:31:56
3
Thomas Wallace
Insanity
07-08-2020 13:20:23
The way how I had thought about approaching this problem is by iterating over the visits and then adding the id to the ride if the name matched, the ride occurred during/after the visit, and the time delta is the smallest difference (using a large initial time delta and then updating the smallest different to that difference). If those conditions are not met, then just keep the same value. With this process in mind, here is my thought process in code:
rides['min_diff'] = pd.to_timedelta(365, unit='day')
rides['id'] = -1
for index, row in visits.iterrows():
rides['id'], rides['min_diff'] = np.where((rides['name'] == row['name']) & (
rides['date'] >= visits['date']) & (
(rides['date'] - row['date']) < rides['min_diff']),
(row['id'], rides['date'] - row['date']),
(rides['id'], rides['min_diff'))
This unfortunately does not execute because of the shapes not matching (as well as trying to assign values across multiple columns, which I am not sure how to do), but this is the general idea. I am not sure how this could be accomplished exactly, so if anyone has a solution, I would appreciate it.
Try with apply() and asof():
df1 = df1.set_index("date").sort_index() #asof requires a sorted index
df2["id"] = df2.apply(lambda x: df1[df1["Name"]==x["Name"]]["id"].asof(x["date"]), axis=1)
>>> df2
Name ride date id
0 John Smith Insanity 2020-07-01 13:53:07 0
1 John Smith Bumper Cars 2020-07-01 16:37:29 0
2 John Smith Tilt-A-Whirl 2020-07-02 08:21:18 0
3 John Smith Insanity 2020-07-22 11:44:32 1
4 Jane Doe Bumper Cars 2020-06-13 14:14:41 2
5 Jane Doe Teacups 2020-06-13 17:31:56 2
6 Thomas Wallace Insanity 2020-07-08 13:20:23 3
I think this does what you need. The ids aren't in the order you specified but they do represent visit ids with the logic you requested.
merged = pd.merge(df1, df2, how="right", left_on=['date', 'name'], right_on=['name', 'ride'])[['name_y', 'ride', 'date_y']]
merged['ymd'] = pd.to_datetime(merged.date_y).apply(lambda x: x.strftime('%Y-%m-%d'))
merged['id'] = merged.groupby(['name_y', 'ymd']).ngroup()
merged.drop('ymd', axis=1, inplace=True)
merged.columns = ['name', 'ride', 'date', 'id']
merged.sort_values(by='id', inplace=True)
print(merged)
OUT:
name ride date id
4 Jane Doe Bumper Cars 06-13-2020 14:14:41 0
5 Jane Doe Teacups 06-13-2020 17:31:56 0
0 John Smith Insanity 07-01-2020 13:53:07 1
1 John Smith Bumper Cars 07-01-2020 16:37:29 1
2 John Smith Tilt-A-Whirl 07-02-2020 08:21:18 2
3 John Smith Insanity 07-22-2020 11:44:32 3
6 Thomas Wallace Insanity 07-08-2020 13:20:23 4
I'm try extract the first two words from a string in dataframe
df["Name"]
Name
Anthony Frank Hawk
John Rodney Mullen
Robert Dean Silva Burnquis
Geoffrey Joseph Rowley
To get index of the second " "(Space) I try this but find return NaN instead return number of characters until second Space.
df["temp"] = df["Name"].str.find(" ")+1
df["temp"] = df["Status"].str.find(" ", start=df["Status"], end=None)
df["temp"]
0 NaN
1 NaN
2 NaN
3 NaN
and the last step is slice those names, I try this code but don't work to.
df["Status"] = df["Status"].str.slice(0,df["temp"])
df["Status"]
0 NaN
1 NaN
2 NaN
3 NaN
expected return
0 Anthony Frank
1 John Rodney
2 Robert Dean
3 Geoffrey Joseph
if you have a more efficient way to do this please let me know!?
df['temp'] = df.Name.str.rpartition().get(0)
df
Output
Name temp
0 Anthony Frank Hawk Anthony Frank
1 John Rodney Mullen John Rodney
2 Robert Dean Silva Burnquis Robert Dean Silva
3 Geoffrey Joseph Rowley Geoffrey Joseph
EDIT
If only first two elements are required in output.
df['temp'] = df.Name.str.split().str[:2].str.join(' ')
df
OR
df['temp'] = df.Name.str.split().apply(lambda x:' '.join(x[:2]))
df
OR
df['temp'] = df.Name.str.split().apply(lambda x:' '.join([x[0], x[1]]))
df
Output
Name temp
0 Anthony Frank Hawk Anthony Frank
1 John Rodney Mullen John Rodney
2 Robert Dean Silva Burnquis Robert Dean
3 Geoffrey Joseph Rowley Geoffrey Joseph
You can use str.index(substring) instead of str.find, it returns the smallest index of the substring(such as " ", empty space) found in the string. Then you can split the string by that index and reapply the above to the second string in the resulting list.
This has to be so simple - but I can't figure it out. I have a "name" column within a DataFrame and I'm trying to reverse the order of ['First Name', 'Middle Name', 'Last Name'] to ['Last Name', 'First Name', 'Middle Name'].
Here is my code:
for i in range(2114):
bb = a['Approved by User'][i].split(" ",2)[2]
aa = a['Approved by User'][i].split(" ",2)[0]
a['Full Name]'] = bb+','+aa
Unfortunately I keep getting IndexError: list index out of range with the current code.
This is what I want:
Old column Name| Jessica Mary Simpson
New column Name| Simpson Jessica Mary
One way to do it is to split the string and joinit later on in a function.
like so:
import pandas as pd
d = {"name": ["Jessica Mary Simpson"]}
df = pd.DataFrame(d)
a = df.name.str.split()
a = a.apply(lambda x: " ".join(x[::-1])).reset_index()
print(a)
output:
index name
0 0 Simpson Mary Jessica
With your shown samples, you could try following.
Let's say following is the df:
fullname
0 Jessica Mary Simpson
1 Ravinder avtar singh
2 John jonny janardan
Here is the code:
df['fullname'].replace(r'^([^ ]*) ([^ ]*) (.*)$', r'\3 \1 \2',regex=True)
OR
df['fullname'].replace(r'^(\S*) (\S*) (.*)$', r'\3 \1 \2',regex=True)
output will be as follows:
0 Simpson Jessica Mary
1 singh Ravinder avtar
2 janardan John jonny
I think problem is in your data, here is your solution in pandas text functions Series.str.split, indexing and Series.str.join:
df['Full Name'] = df['Approved by User'].str.split(n=2).str[::-1].str.join(' ')
print (df)
Approved by User Full Name
0 Jessica Mary Simpson Simpson Mary Jessica
1 John Doe Doe John
2 Mary Mary
I'm trying to replace column "Names" by a new variable "Gender" based on the first letters that we find in column name.
INPUT:
df['Name'].value_counts()
OUTPUT:
Mr. Gordon Hemmings 1
Miss Jane Wilkins 1
Mrs. Audrey North 1
Mrs. Wanda Sharp 1
Mr. Victor Hemmings 1
..
Miss Heather Abraham 1
Mrs. Kylie Hart 1
Mr. Ian Langdon 1
Mr. Gordon Watson 1
Miss Irene Vance 1
Name: Name, Length: 4999, dtype: int64
Now, see the Miss, Mrs., and Miss? The first question that comes to mind is: how many different words there are?
INPUT
df.Name.str.split().str[0].value_counts(dropna=False)
Mr. 3351
Mrs. 937
Miss 711
NaN 1
Name: Name, dtype: int64
Now I'm trying to:
#Replace missing value
df['Name'].fillna('Mr.', inplace=True)
# Create Column Gender
df['Gender'] = df['Name']
for i in range(0, df[0]):
A = df['Name'].values[i][0:3]=="Mr."
df['Gender'].values[i] = A
df.loc[df['Gender']==True, 'Gender']="Male"
df.loc[df['Gender']==False, 'Gender']="Female"
del df['Name'] #Delete column 'Name'
df
But I'm missing something since I get the following error:
KeyError: 0
The KeyError is because you don't have a column called 0. However, I would ditch that code and try something more efficient.
You can use np.where with str.contains to search for names with Mr. after using fillna(). Then, just drop the Name column.:
df['Name'] = df['Name'].fillna('Mr.')
df['Gender'] = np.where(df['Name'].str.contains('Mr\.'), 'Male', 'Female')
df = df.drop('Name', axis=1)
df
Full example:
df = pd.DataFrame({'Name': {0: 'Mr. Gordon Hemmings',
1: 'Miss Jane Wilkins',
2: 'Mrs. Audrey North',
3: 'Mrs. Wanda Sharp',
4: 'Mr. Victor Hemmings'},
'Value': {0: 1, 1: 1, 2: 1, 3: 1, 4: 1}})
print(df)
df['Name'] = df['Name'].fillna('Mr.')
df['Gender'] = np.where(df['Name'].str.contains('Mr\.'), 'Male', 'Female')
df = df.drop('Name', axis=1)
print('\n')
print(df)
Name Value
0 Mr. Gordon Hemmings 1
1 Miss Jane Wilkins 1
2 Mrs. Audrey North 1
3 Mrs. Wanda Sharp 1
4 Mr. Victor Hemmings 1
Value Gender
0 1 Male
1 1 Female
2 1 Female
3 1 Female
4 1 Male
Background
I have a toy df
import pandas as pd
df = pd.DataFrame({'Text' : ['Jon J Mmith is Here',
'Mary Lisa Hder found here',
'Jane A Doe is also here',
'Tom T Tcker is here too'],
'P_ID': [1,2,3,4],
'P_Name' : ['MMITH, JON J', 'HDER, MARY LISA', 'DOE, JANE A', 'TCKER, TOM T'],
'N_ID' : ['A1', 'A2', 'A3', 'A4']
})
#rearrange columns
df = df[['Text','N_ID', 'P_ID', 'P_Name']]
df
Text N_ID P_ID P_Name
0 Jon J Mmith is Here A1 1 MMITH, JON J
1 Mary Lisa Hder found here A2 2 HDER, MARY LISA
2 Jane A Doe is also here A3 3 DOE, JANE A
3 Tom T Tcker is here to A4 4 TCKER, TOM T
Goal
1) Change the P_Name column from df into a format that looks like my desired output; that is, change the current format (e.g.MMITH, JON J) to a format (e.g. Mmith, Jon J) where the first and last names and middle letter all start with a capital letter
2) Create this in a new column P_Name_New
Desired Output
Text N_ID P_ID P_Name P_Name_New
0 Jon J Mmith is Here A1 1 MMITH, JON J Mmith, Jon J
1 Mary Lisa Hder found here A2 2 HDER, MARY LISA Hder, Mary Lisa
2 Jane A Doe is also here A3 3 DOE, JANE A Doe, Jane A
3 Tom T Tcker is here too A4 4 TCKER, TOM T Tcker, Tom T
Question
How do I achieve my desired goal?
Simply with str.title() function:
In [98]: df['P_Name_New'] = df['P_Name'].str.title()
In [99]: df
Out[99]:
Text N_ID P_ID P_Name P_Name_New
0 Jon J Smith is Here A1 1 SMITH, JON J Smith, Jon J
1 Mary Lisa Rider found here A2 2 RIDER, MARY LISA Rider, Mary Lisa
2 Jane A Doe is also here A3 3 DOE, JANE A Doe, Jane A
3 Tom T Tucker is here too A4 4 TUCKER, TOM T Tucker, Tom T