Develop Reference

Using melt() in Pandas

Trying to melt the following data such that the result is 4 rows - one for "John-1" containing his before data, one for "John-2" containing his after data, one for "Kelly-1" containing her before data, and one for "Kelly-2" containing her after data. And the columns would be "Name", "Weight", and "Height". Can this be done solely with the melt function?
df = pd.DataFrame({'Name': ['John', 'Kelly'],
'Weight Before': [200, 175],
'Weight After': [195, 165],
'Height Before': [6, 5],
'Height After': [7, 6]})

Use pandas.wide_to_long function as shown below:
pd.wide_to_long(df, ['Weight', 'Height'], 'Name', 'grp', ' ', '\\w+').reset_index()
Name grp Weight Height
0 John Before 200 6
1 Kelly Before 175 5
2 John After 195 7
3 Kelly After 165 6
or you could also use pivot_longer from pyjanitor as follows:
import janitor
df.pivot_longer('Name', names_to = ['.value', 'grp'], names_sep = ' ')
Name grp Weight Height
0 John Before 200 6
1 Kelly Before 175 5
2 John After 195 7
3 Kelly After 165 6

Pandas Series and Nan Values for mismatched values

I have these two dictionaries,
dico = {'Name': ['Arthur','Henri','Lisiane','Patrice','Zadig','Sacha'],
"Age": ["20","18","62","73",'21','20'],
"Studies": ['Economics','Maths','Psychology','Medical','Cinema','CS']
}
dico2 = {'Surname': ['Arthur1','Henri2','Lisiane3','Patrice4']}
dico = pd.DataFrame.from_dict(dico)
dico2 = pd.DataFrame.from_dict(dico2)
in which I would like to match then append the Surname column with the Name column, to finally append it to dico, for a following output:
Name Surname Age Studies
0 Arthur Arthur1 20 Economics
1 Henri Henri2 18 Maths
2 Lisiane Lisiane3 62 Psychology
3 Patrice Nan 73 Medical
4 Zadig Nan 21 Cinema
5 Sacha Nan 20 CS
and ultimately delete the rows for which Surname is Nan
Name Surname Age Studies
0 Arthur Arthur1 20 Economics
1 Henri Henri2 18 Maths
2 Lisiane Lisiane3 62 Psychology
map_list = []
for name in dico['Name']:
best_ratio = None
for idx, surname in enumerate(dico2['Surname']):
if best_ratio == None:
best_ratio = fuzz.ratio(name, surname)
best_idx = 0
else:
ratio = fuzz.ratio(name, surname)
if ratio > best_ratio:
best_ratio = ratio
best_idx = idx
map_list.append(dico2['Surname'][best_idx]) # obtain surname
dico['Surname'] = pd.Series(map_list) # add column
dico = dico[["Name", "Surname", "Age", "Studies"]] # reorder columns
#if the surname is not a great match, print "Nan"
dico = dico.drop(dico[dico.Surname == "NaN"].index)
but when I print(dico), the output is as follows:
Name Surname Age Studies
0 Arthur Arthur1 20 Economics
1 Henri Henri2 18 Maths
2 Lisiane Lisiane3 62 Psychology
3 Patrice Patrice4 73 Medical
4 Zadig Patrice4 21 Cinema
5 Sacha Patrice4 20 CS
I don't see why after the Patrice row, there's a mismatch, while I want it to be "Nan".

Lets try pd.Multiindex.from_product to create combinations and then assign a score with zip and fuzz.ratio and some filtering to create our dict, then we can use series.map and df.dropna:
from fuzzywuzzy import fuzz
comb = pd.MultiIndex.from_product((dico['Name'],dico2['Surname']))
scores = comb.map(lambda x: fuzz.ratio(*x)) #or fuzz.partial_ratio(*x)
d = dict(a for a,b in zip(comb,scores) if b>90) #change threshold
out = dico.assign(SurName=dico['Name'].map(d)).dropna(subset=['SurName'])
print(out)
Name Age Studies SurName
0 Arthur 20 Economics Arthur1
1 Henri 18 Maths Henri2
2 Lisiane 62 Psychology Lisiane3
3 Patrice 73 Medical Patrice4

You could do the following thing. Define the function:
def fuzzy_merge(df_1, df_2, key1, key2, threshold=90, limit=2):
s = df_2[key2].tolist()
m = df_1[key1].apply(lambda x: process.extract(x, s, limit=limit))
df_1['Surname'] = m
m2 = df_1['Surname'].apply(lambda x: ', '.join([i[0] for i in x if i[1] >= threshold]))
df_1['Surname'] = m2
return df_1
and run
from fuzzywuzzy import fuzz
from fuzzywuzzy import process
df = fuzzy_merge(dico, dico2, 'Name', 'Surname',threshold=90, limit=2)
This returns:
Name Age Studies Surname
0 Arthur 20 Economics Arthur1
1 Henri 18 Maths Henri2
2 Lisiane 62 Psychology Lisiane3
3 Patrice 73 Medical Patrice4
4 Zadig 21 Cinema
5 Sacha 20 CS

Faster way to query & compute in Pandas [duplicate]

This question already has answers here:
Pandas Merging 101
(8 answers)
Closed 2 years ago.
I have two dataframes in Pandas. What I want achieve is, grab every 'Name' from DF1 and get the corresponding 'City' and 'State' present in DF2.
For example, 'Dwight' from DF1 should return corresponding values 'Miami' and 'Florida' from DF2.
DF1
Name Age Student
0 Dwight 20 Yes
1 Michael 30 No
2 Pam 55 No
. . . .
70000 Jim 27 Yes
DF1 has approx 70,000 rows with 3 columns
Second Dataframe, DF2 has approx 320,000 rows.
Name City State
0 Dwight Miami Florida
1 Michael Scranton Pennsylvania
2 Pam Austin Texas
. . . . .
325082 Jim Scranton Pennsylvania
Currently I have two functions, which return the values of 'City' and 'State' using a filter.
def read_city(id):
filt = (df2['Name'] == id)
if filt.any():
field = (df2[filt]['City'].values[0])
else:
field = ""
return field
def read_state(id):
filt = (df2['Name'] == id)
if filt.any():
field = (df2[filt]['State'].values[0])
else:
field = ""
return field
I am using the apply function to process all the values.
df['city_list'] = df['Name'].apply(read_city)
df['State_list'] = df['Name'].apply(read_state)
The result takes a long time to compute in the above way. It roughly takes me around 18 minutes to get back the df['city_list'] and df['State_list'].
Is there a faster to compute this ? Since I am completely new to pandas, I would like to know if there is a efficient way to compute this ?

I believe you can do a map:
s = df2.groupby('name')[['City','State']].agg(list)
df['city_list'] = df['Name'].map(s['City'])
df['State_list'] = df['Name'].map(s['State'])
Or a left merge after you got s:
df = df.merge(s.add_suffix('_list'), left_on='Name', right_index=True, how='left')

I think you can do something like this:
# Dataframe DF1 (dummy data)
DF1 = pd.DataFrame(columns=['Name', 'Age', 'Student'], data=[['Dwight', 20, 'Yes'], ['Michael', 30, 'No'], ['Pam', 55, 'No'], ['Jim', 27, 'Yes']])
print("DataFrame DF1")
print(DF1)
# Dataframe DF2 (dummy data)
DF2 = pd.DataFrame(columns=['Name', 'City', 'State'], data=[['Dwight', 'Miami', 'Florida'], ['Michael', 'Scranton', 'Pennsylvania'], ['Pam', 'Austin', 'Texas'], ['Jim', 'Scranton', 'Pennsylvania']])
print("DataFrame DF2")
print(DF2)
# You do a merge on 'Name' column and then, you change the name of columns 'City' and 'State'
df = pd.merge(DF1, DF2, on=['Name']).rename(columns={'City': 'city_list', 'State': 'State_list'})
print("DataFrame final")
print(df)
Output:
DataFrame DF1
Name Age Student
0 Dwight 20 Yes
1 Michael 30 No
2 Pam 55 No
3 Jim 27 Yes
DataFrame DF2
Name City State
0 Dwight Miami Florida
1 Michael Scranton Pennsylvania
2 Pam Austin Texas
3 Jim Scranton Pennsylvania
DataFrame final
Name Age Student city_list State_list
0 Dwight 20 Yes Miami Florida
1 Michael 30 No Scranton Pennsylvania
2 Pam 55 No Austin Texas
3 Jim 27 Yes Scranton Pennsylvania

add a row at top in pandas dataframe [duplicate]

This question already has answers here:
Insert a row to pandas dataframe
(18 answers)
Closed 4 years ago.
Below is my dataframe
import pandas as pd
df = pd.DataFrame({'name': ['jon','sam','jane','bob'],
'age': [30,25,18,26],
'sex':['male','male','female','male']})
age name sex
0 30 jon male
1 25 sam male
2 18 jane female
3 26 bob male
I want to insert a new row at the first position
name: dean, age: 45, sex: male
age name sex
0 45 dean male
1 30 jon male
2 25 sam male
3 18 jane female
4 26 bob male
What is the best way to do this in pandas?

Probably this is not the most efficient way but:
df.loc[-1] = ['45', 'Dean', 'male'] # adding a row
df.index = df.index + 1 # shifting index
df.sort_index(inplace=True)
Output:
age name sex
0 45 Dean male
1 30 jon male
2 25 sam male
3 18 jane female
4 26 bob male

If it's going to be a frequent operation, then it makes sense (in terms of performance) to gather the data into a list first and then use pd.concat([], ignore_index=True) (similar to #Serenity's solution):
Demo:
data = []
# always inserting new rows at the first position - last row will be always on top
data.insert(0, {'name': 'dean', 'age': 45, 'sex': 'male'})
data.insert(0, {'name': 'joe', 'age': 33, 'sex': 'male'})
#...
pd.concat([pd.DataFrame(data), df], ignore_index=True)
In [56]: pd.concat([pd.DataFrame(data), df], ignore_index=True)
Out[56]:
age name sex
0 33 joe male
1 45 dean male
2 30 jon male
3 25 sam male
4 18 jane female
5 26 bob male
PS I wouldn't call .append(), pd.concat(), .sort_index() too frequently (for each single row) as it's pretty expensive. So the idea is to do it in chunks...

#edyvedy13's solution worked great for me. However it needs to be updated for the deprecation of pandas' sort method - now replaced with sort_index.
df.loc[-1] = ['45', 'Dean', 'male'] # adding a row
df.index = df.index + 1 # shifting index
df = df.sort_index() # sorting by index

Use pandas.concat and reindex new dataframe:
import pandas as pd
df = pd.DataFrame({'name': ['jon','sam','jane','bob'],
'age': [30,25,18,26],
'sex':['male','male','female','male']})
# new line
line = pd.DataFrame({'name': 'dean', 'age': 45, 'sex': 'male'}, index=[0])
# concatenate two dataframe
df2 = pd.concat([line,df.ix[:]]).reset_index(drop=True)
print (df2)
Output:
age name sex
0 45 dean male
1 30 jon male
2 25 sam male
3 18 jane female
4 26 bob male

import pandas as pd
df = pd.DataFrame({'name': ['jon','sam','jane','bob'],
'age': [30,25,18,26],
'sex': ['male','male','female','male']})
df1 = pd.DataFrame({'name': ['dean'], 'age': [45], 'sex':['male']})
df1 = df1.append(df)
df1 = df1.reset_index(drop=True)
That works

This will work for me.
>>> import pandas as pd
>>> df = pd.DataFrame({'name': ['jon','sam','jane','bob'],
... 'age': [30,25,18,26],
... 'sex':['male','male','female','male']}) >>> df
age name sex
0 30 jon male
1 25 sam male
2 18 jane female
3 26 bob male
>>> df.loc['a']=[45,'dean','male']
>>> df
age name sex
0 30 jon male
1 25 sam male
2 18 jane female
3 26 bob male
a 45 dean male
>>> newIndex=['a']+[ind for ind in df.index if ind!='a']
>>> df=df.reindex(index=newIndex)
>>> df
age name sex
a 45 dean male
0 30 jon male
1 25 sam male
2 18 jane female
3 26 bob male

Splitting multiple pipe delimited values in multiple columns of a comma delimited CSV and mapping them to each other

I have a csv with comma delimiters that has multiple values in a column that are delimited by a pipe and I need to map them to another column with multiple pipe delimited values and then give them their own row along with data in the original row that doesn't have multiple values. My CSV looks like this (with commas between the categories):
row name city amount
1 frank | john | dave toronto | new york | anaheim 10
2 george | joe | fred fresno | kansas city | reno 20
I need it to look like this:
row name city amount
1 frank toronto 10
2 john new york 10
3 dave anaheim 10
4 george fresno 20
5 joe kansas city 20
6 fred reno 20

Maybe not the nicest but working solution:
(works with no piped lines and for different pipe-length)
df = pd.read_csv('<your_data>.csv')
str_split = ' | '
# Calculate maximum length of piped (' | ') values
df['max_len'] = df[['name', 'city']].apply(lambda x: max(len(x[0].split(str_split)),
len(x[0].split(str_split))), axis=1)
max_len = df['max_len'].max()
# Split '|' piped cell values into columns (needed at unpivot step)
# Create as many new 'name_<x>' & 'city_<x>' columns as 'max_len'
df[['name_{}'.format(i) for i in range(max_len)]] = df['name'].apply(lambda x: \
pd.Series(x.split(str_split)))
df[['city_{}'.format(i) for i in range(max_len)]] = df['city'].apply(lambda x: \
pd.Series(x.split(str_split)))
# Unpivot 'name_<x>' & 'city_<x>' columns into rows
df_pv_name = pd.melt(df, value_vars=['name_{}'.format(i) for i in range(max_len)],
id_vars=['amount'])
df_pv_city = pd.melt(df, value_vars=['city_{}'.format(i) for i in range(max_len)],
id_vars=['amount'])
# Rename upivoted columns (these are the final columns)
df_pv_name = df_pv_name.rename(columns={'value':'name'})
df_pv_city = df_pv_city.rename(columns={'value':'city'})
# Rename 'city_<x>' values (rows) to be 'key' for join (merge)
df_pv_city['variable'] = df_pv_city['variable'].map({'city_{}'.format(i):'name_{}'\
.format(i) for i in range(max_len)})
# Join unpivoted 'name' & 'city' dataframes
df_res = df_pv_name.merge(df_pv_city, on=['variable', 'amount'])
# Drop 'variable' column and NULL rows if you have not equal pipe-length in original rows
# If you want to drop any NULL rows then replace 'all' to 'any'
df_res = df_res.drop(['variable'], axis=1).dropna(subset=['name', 'city'], how='all',
axis=0).reset_index(drop=True)
The result is:
amount name city
0 10 frank toronto
1 20 george fresno
2 10 john new york
3 20 joe kansas city
4 10 dave anaheim
5 20 fred reno
Another test input:
name city amount
0 frank | john | dave | joe | bill toronto | new york | anaheim | los angeles | caracas 10
1 george | joe | fred fresno | kansas city 20
2 danny miami 30
Result of this test (if you don't want NaN rows then replace how='all' to how='any' in the code at merging):
amount name city
0 10 frank toronto
1 20 george fresno
2 30 danny miami
3 10 john new york
4 20 joe kansas city
5 10 dave anaheim
6 20 fred NaN
7 10 joe los angeles
8 10 bill caracas

Given a row:
['1','frank|joe|dave', 'toronto|new york|anaheim', '20']
you can use
itertools.izip_longest(*[value.split('|') for value in row])
on it to obtain following structure:
[('1', 'frank', 'toronto', '20'),
(None, 'joe', 'new york', None),
(None, 'dave', 'anaheim', None)]
Here we want to replace all None values with last seen value in corresponding column. Can be done when looping over result.
So given a TSV already splitted by tabs following code should do the trick:
import itertools
def flatten_tsv(lines):
result = []
for line in lines:
flat_lines = itertools.izip_longest(*[value.split('|') for value in line])
for flat_line in flat_lines:
result.append([result[-1][i] if v is None else v
for i, v in enumerate(flat_line)])
return result