相当于粘贴 R 到 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 that I have this dataframe:

   categorie titre tarifMin  lieu  long   lat   img dateSortie
1      zoo,  Aquar      0.0 Aquar 2.385 48.89 ilo,0           
2      zoo,  Aquar      4.5 Aquar 2.408 48.83 ilo,0           
6      lieu  Jardi      0.0 Jardi 2.320 48.86 ilo,0           
7      lieu  Bois       0.0 Bois  2.455 48.82 ilo,0           
13     espac Canal      0.0 Canal 2.366 48.87 ilo,0           
14     espac Canal     -1.0 Canal 2.384 48.89 ilo,0           
15     parc  Le Ma     20.0 Le Ma 2.353 48.87 ilo,0 

I want to create a new column which uses another column in a dataframe and some text. With R, I do:

> y$thecolThatIWant=ifelse(y$tarifMin!=-1,
+                             paste("Evenement permanent  -->",y$categorie,
+                                   y$titre,"C  partir de",y$tarifMin,"€uros"),
+                             paste("Evenement permanent  -->",y$categorie,
+                                   y$titre,"sans prix indique"))

And the result is:

> y
   categorie titre tarifMin  lieu  long   lat   img dateSortie
1      zoo,  Aquar      0.0 Aquar 2.385 48.89 ilo,0           
2      zoo,  Aquar      4.5 Aquar 2.408 48.83 ilo,0           
6      lieu  Jardi      0.0 Jardi 2.320 48.86 ilo,0           
7      lieu  Bois       0.0 Bois  2.455 48.82 ilo,0           
13     espac Canal      0.0 Canal 2.366 48.87 ilo,0           
14     espac Canal     -1.0 Canal 2.384 48.89 ilo,0           
15     parc  Le Ma     20.0 Le Ma 2.353 48.87 ilo,0           
                                                thecolThatIWant
1  Evenement permanent  --> zoo,  Aquar C  partir de  0.0 €uros
2  Evenement permanent  --> zoo,  Aquar C  partir de  4.5 €uros
6  Evenement permanent  --> lieu  Jardi C  partir de  0.0 €uros
7  Evenement permanent  --> lieu  Bois  C  partir de  0.0 €uros
13 Evenement permanent  --> espac Canal C  partir de  0.0 €uros
14 Evenement permanent  --> espac Canal C  partir de -1.0 €uros
15 Evenement permanent  --> parc  Le Ma C  partir de 20.0 €uros

My question is: How can I do the same thing in Python Pandas or some other module?

What I've tried so far: Well, I'm a very new user. So sorry for my mistake. I try to replicate the example in Python and we suppose that I get something like this

table=pd.read_csv("y.csv",sep=",")
tt= table.loc[:,['categorie','titre','tarifMin','long','lat','lieu']]
table
ategorie    titre   tarifMin    long    lat     lieu
0   zoo,    Aquar   0.0     2.385   48.89   Aquar
1   zoo,    Aquar   4.5     2.408   48.83   Aquar
2   lieu    Jardi   0.0     2.320   48.86   Jardi
3   lieu    Bois    0.0     2.455   48.82   Bois
4   espac   Canal   0.0     2.366   48.87   Canal
5   espac   Canal   -1.0    2.384   48.89   Canal
6   parc    Le Ma   20.0    2.353   48.87   Le Ma

I tried this basically

sc="Even permanent -->" + " "+ tt.titre+" "+tt.lieu
tt['theColThatIWant'] = sc
tt

And I got this

    categorie   titre   tarifMin    long    lat     lieu    theColThatIWant
0   zoo,    Aquar   0.0     2.385   48.89   Aquar   Even permanent --> Aquar Aquar
1   zoo,    Aquar   4.5     2.408   48.83   Aquar   Even permanent --> Aquar Aquar
2   lieu    Jardi   0.0     2.320   48.86   Jardi   Even permanent --> Jardi Jardi
3   lieu    Bois    0.0     2.455   48.82   Bois    Even permanent --> Bois Bois
4   espac   Canal   0.0     2.366   48.87   Canal   Even permanent --> Canal Canal
5   espac   Canal   -1.0    2.384   48.89   Canal   Even permanent --> Canal Canal
6   parc    Le Ma   20.0    2.353   48.87   Le Ma   Even permanent --> Le Ma Le Ma

Now, I suppose that I have to loop with condition if there is no vectorize like in R?

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

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 3

For this particular case, the paste operator in R is closest to Python's format which was added in Python 2.6. It's newer and somewhat more flexible than the older % operator.

For a purely Python-ic answer without using numpy or pandas, here is one way to do it using your original data in the form of a list of lists (this could also have been done as a list of dict, but that seemed more cluttered to me).

# -*- coding: utf-8 -*-
names=['categorie','titre','tarifMin','lieu','long','lat','img','dateSortie']

records=[[
    'zoo',   'Aquar',     0.0,'Aquar',2.385,48.89,'ilo',0],[
    'zoo',   'Aquar',     4.5,'Aquar',2.408,48.83,'ilo',0],[
    'lieu',  'Jardi',     0.0,'Jardi',2.320,48.86,'ilo',0],[
    'lieu',  'Bois',      0.0,'Bois', 2.455,48.82,'ilo',0],[
    'espac', 'Canal',     0.0,'Canal',2.366,48.87,'ilo',0],[
    'espac', 'Canal',    -1.0,'Canal',2.384,48.89,'ilo',0],[
    'parc',  'Le Ma',    20.0,'Le Ma', 2.353,48.87,'ilo',0] ]

def prix(p):
    if (p != -1):
        return 'C  partir de {} €uros'.format(p)
    return 'sans prix indique'

def msg(a):
    return 'Evenement permanent  --> {}, {} {}'.format(a[0],a[1],prix(a[2]))

[m.append(msg(m)) for m in records]

from pprint import pprint

pprint(records)

The result is this:

[['zoo',
  'Aquar',
  0.0,
  'Aquar',
  2.385,
  48.89,
  'ilo',
  0,
  'Evenement permanent  --> zoo, Aquar C  partir de 0.0 \xe2\x82\xacuros'],
 ['zoo',
  'Aquar',
  4.5,
  'Aquar',
  2.408,
  48.83,
  'ilo',
  0,
  'Evenement permanent  --> zoo, Aquar C  partir de 4.5 \xe2\x82\xacuros'],
 ['lieu',
  'Jardi',
  0.0,
  'Jardi',
  2.32,
  48.86,
  'ilo',
  0,
  'Evenement permanent  --> lieu, Jardi C  partir de 0.0 \xe2\x82\xacuros'],
 ['lieu',
  'Bois',
  0.0,
  'Bois',
  2.455,
  48.82,
  'ilo',
  0,
  'Evenement permanent  --> lieu, Bois C  partir de 0.0 \xe2\x82\xacuros'],
 ['espac',
  'Canal',
  0.0,
  'Canal',
  2.366,
  48.87,
  'ilo',
  0,
  'Evenement permanent  --> espac, Canal C  partir de 0.0 \xe2\x82\xacuros'],
 ['espac',
  'Canal',
  -1.0,
  'Canal',
  2.384,
  48.89,
  'ilo',
  0,
  'Evenement permanent  --> espac, Canal sans prix indique'],
 ['parc',
  'Le Ma',
  20.0,
  'Le Ma',
  2.353,
  48.87,
  'ilo',
  0,
  'Evenement permanent  --> parc, Le Ma C  partir de 20.0 \xe2\x82\xacuros']]

Note that although I've defined a list names it isn't actually used. One could define a dictionary with the names of the titles as the key and the field number (starting from 0) as the value, but I didn't bother with this to try to keep the example simple.

The functions prix and msg are fairly simple. The only tricky portion is the list comprehension [m.append(msg(m)) for m in records] which iterates through all of the records, and modifies each to append your new field, created via a call to msg .

Answer 4

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 5

1. You can try pandas.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 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

Answer 7

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. Eg, 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 8

This is simple example how to achive that (If I'am not worng what do you want to do):

import numpy as np
import pandas as pd

dates = pd.date_range('20130101',periods=6)
df = pd.DataFrame(np.random.randn(6,4),index=dates,columns=list('ABCD'))
for row in df.itertuples():
    index, A, B, C, D = row
    print '%s Evenement permanent  --> %s , next data %s' % (index, A, B)

Output:

>>>df
                   A         B         C         D
2013-01-01 -0.400550 -0.204032 -0.954237  0.019025
2013-01-02  0.509040 -0.611699  1.065862  0.034486
2013-01-03  0.366230  0.805068 -0.144129 -0.912942
2013-01-04  1.381278 -1.783794  0.835435 -0.140371
2013-01-05  1.140866  2.755003 -0.940519 -2.425671
2013-01-06 -0.610569 -0.282952  0.111293 -0.108521

This what loop for print: 2013-01-01 00:00:00 Evenement permanent --> -0.400550121168 , next data -0.204032344442

2013-01-02 00:00:00 Evenement permanent  --> 0.509040318928 , next data -0.611698560541

2013-01-03 00:00:00 Evenement permanent  --> 0.366230438863 , next data 0.805067758304

2013-01-04 00:00:00 Evenement permanent  --> 1.38127775713 , next data -1.78379439485

2013-01-05 00:00:00 Evenement permanent  --> 1.14086631509 , next data 2.75500268167

2013-01-06 00:00:00 Evenement permanent  --> -0.610568516983 , next data -0.282952162792

Answer 9

There is actually a very easy way. You just convert your variable to a string . For instance, try to run this:

a = 1; b = "you are number " + str(a); b