Equivalent of Paste R to Python

Question

I am a new python afficionado. For R users, there is one function : paste that helps to concatenate two or more variables in a dataframe. It\'s very useful. For example Suppose

Answer 1

This very much works like Paste command in R: R code:

 words = c("Here", "I","want","to","concatenate","words","using","pipe","delimeter")
 paste(words,collapse="|")

[1]

"Here|I|want|to|concatenate|words|using|pipe|delimeter"

Python:

words = ["Here", "I","want","to","concatenate","words","using","pipe","delimeter"]
"|".join(words)

Result:

'Here|I|want|to|concatenate|words|using|pipe|delimeter'

Answer 2

1. You can trypandas.Series.str.cat

   import pandas as pd
   def paste0(ss,sep=None,na_rep=None,):
       '''Analogy to R paste0'''
       ss = [pd.Series(s) for s in ss]
       ss = [s.astype(str) for s in ss]
       s = ss[0]
       res = s.str.cat(ss[1:],sep=sep,na_rep=na_rep)
       return res
   
   pasteA=paste0
   

2. Or just sep.join()

   #

   def paste0(ss,sep=None,na_rep=None, 
       castF=unicode, ##### many languages dont work well with str
   ):
       if sep is None:
           sep=''
       res = [castF(sep).join(castF(s) for s in x) for x in zip(*ss)]
       return res
   pasteB = paste0
   
   
   %timeit pasteA([range(1000),range(1000,0,-1)],sep='_')
   # 100 loops, best of 3: 7.11 ms per loop
   %timeit pasteB([range(1000),range(1000,0,-1)],sep='_')
   # 100 loops, best of 3: 2.24 ms per loop
   

3. I have used itertools to mimic recycling

   import itertools
   def paste0(ss,sep=None,na_rep=None,castF=unicode):
       '''Analogy to R paste0
       '''
       if sep is None:
           sep=u''
       L = max([len(e) for e in ss])
       it = itertools.izip(*[itertools.cycle(e) for e in ss])
       res = [castF(sep).join(castF(s) for s in next(it) ) for i in range(L)]
       # res = pd.Series(res)
       return res
   

4. patsy might be relevant (not an experienced user myself.)

Answer 3

Here's a simple implementation that works on lists, and probably other iterables. Warning: it's only been lightly tested, and only in Python 3.5+:

from functools import reduce

def _reduce_concat(x, sep=""):
    return reduce(lambda x, y: str(x) + sep + str(y), x)
        
def paste(*lists, sep=" ", collapse=None):
    result = map(lambda x: _reduce_concat(x, sep=sep), zip(*lists))
    if collapse is not None:
        return _reduce_concat(result, sep=collapse)
    return list(result)

assert paste([1,2,3], [11,12,13], sep=',') == ['1,11', '2,12', '3,13']
assert paste([1,2,3], [11,12,13], sep=',', collapse=";") == '1,11;2,12;3,13'

You can also have some more fun and replicate other functions like paste0:

from functools import partial

paste0 = partial(paste, sep="")

Edit: here's a Repl.it project with type-annotated versions of this code.

Answer 4

If you want to just paste two string columns together, you can simplify @shouldsee's answer because you don't need to create the function. E.g., in my case:

df['newcol'] = df['id_part_one'].str.cat(df['id_part_two'], sep='_')

It might be required for both Series to be of dtype object in order to this (I haven't verified).

Answer 5

my anwser is loosely based on original question, was edited from answer by woles. I would like to illustrate the points:

• paste is % operator in python
• using apply you can make new value and assign it to new column

for R folks: there is no ifelse in direct form (but there are ways to nicely replace it).

import numpy as np
import pandas as pd

dates = pd.date_range('20140412',periods=7)
df = pd.DataFrame(np.random.randn(7,4),index=dates,columns=list('ABCD'))
df['categorie'] = ['z', 'z', 'l', 'l', 'e', 'e', 'p']

def apply_to_row(x):
    ret = "this is the value i want: %f" % x['A']
    if x['B'] > 0:
        ret = "no, this one is better: %f" % x['C']
    return ret

df['theColumnIWant'] = df.apply(apply_to_row, axis = 1)
print df

Answer 6

Let's try things with apply.

df.apply( lambda x: str( x.loc[ desired_col ] ) + "pasting?" , axis = 1 )

you will recevied things similar like paste