等价于将 R 粘贴到 Python
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Equivalent of Paste R to Python
提问by GjT
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 :
我是一个新的 python 爱好者。对于 R 用户，有一个功能： paste 有助于连接数据帧中的两个或多个变量。这是非常有用的。例如假设我有这个数据框：
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 :
我想创建一个新列，它使用数据框中的另一列和一些文本。使用 R，我这样做：
> 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 :
结果是：
> 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?
我的问题是：如何在 Python Pandas 或其他模块中做同样的事情？
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
到目前为止我尝试过的：嗯，我是一个非常新的用户。很抱歉我的错误。我尝试在 Python 中复制这个例子，我们假设我得到了这样的东西
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
我基本上试过这个
sc="Even permanent -->" + " "+ tt.titre+" "+tt.lieu
tt['theColThatIWant'] = sc
tt
And I got this
我得到了这个
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?
现在，我想如果没有像 R 中那样的矢量化，我必须用条件循环？
回答by Micha?
This is simple example how to achive that (If I'am not worng what do you want to do):
这是如何实现这一目标的简单示例（如果我不知道你想做什么）：
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:
输出：
>>>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
这是什么循环打印： 2013-01-01 00:00:00 Evenement 永久 --> -0.400550121168 ，下一个数据 -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
回答by lowtech
my anwser is loosely based on original question, was edited from answer by woles. I would like to illustrate the points:
我的回答大致基于原始问题，是根据 woles 的回答编辑的。我想说明以下几点：
- paste is % operator in python
- using apply you can make new value and assign it to new column
- 粘贴是python中的%运算符
- 使用 apply 您可以创建新值并将其分配给新列
for R folks: there is no ifelse in direct form (but there are ways to nicely replace it).
对于 R 人员：没有直接形式的 ifelse（但有一些方法可以很好地替换它）。
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
回答by Edward
For this particular case, the pasteoperator in Ris closest to Python's formatwhich was added in Python 2.6. It's newer and somewhat more flexible than the older %operator.
对于这种特殊情况，paste操作符 inR最接近formatPython 2.6 中添加的 Python操作符。它比旧的%运算符更新，并且更灵活。
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).
对于不使用 numpy 或 Pandas 的纯 Python ic 答案，这是使用列表列表形式的原始数据的一种方法（这也可以作为 dict 列表完成，但这似乎更混乱对我来说）。
# -*- 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:
结果是这样的：
[['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 namesit 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.
请注意，虽然我定义了一个列表，names但实际上并没有使用它。可以定义一个字典，以标题的名称作为键，以字段编号（从 0 开始）作为值，但我没有费心去尝试使示例保持简单。
The functions prixand msgare 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.
功能prix和msg是相当简单的。唯一棘手的部分是列表理解[m.append(msg(m)) for m in records]，它遍历所有记录，并修改每个记录以附加您的新字段，该字段是通过调用msg.
回答by shadowtalker
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:
这是一个适用于列表的简单实现，可能还有其他可迭代对象。警告：它只经过轻微测试，并且仅在 Python 3.5 中：
import functools
def reduce_concat(x, sep=""):
return functools.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)
print(paste([1,2,3], [11,12,13], sep=','))
print(paste([1,2,3], [11,12,13], sep=',', collapse=";"))
# ['1,11', '2,12', '3,13']
# '1,11;2,12;3,13'
You can also have some more fun and replicate other functions like paste0:
您还可以享受更多乐趣并复制其他功能，例如paste0：
paste0 = functools.partial(paste, sep="")
回答by SAHIL BHANGE
This very much works like Paste command in R: R code:
这非常类似于 R: R 代码中的 Paste 命令：
words = c("Here", "I","want","to","concatenate","words","using","pipe","delimeter")
paste(words,collapse="|")
[1]
[1]
"Here|I|want|to|concatenate|words|using|pipe|delimeter"
“这里|我|想要|要|连接|单词|使用|管道|分隔符”
Python:
Python：
words = ["Here", "I","want","to","concatenate","words","using","pipe","delimeter"]
"|".join(words)
Result:
结果：
'Here|I|want|to|concatenate|words|using|pipe|delimeter'
'这里|我|想要|要|连接|单词|使用|管道|分隔符'
回答by u7137467
Let's try things with apply.
让我们用apply试试。
df.apply( lambda x: str( x.loc[ desired_col ] ) + "pasting?" , axis = 1 )
you will recevied things similar like paste
你会收到类似粘贴的东西
回答by shouldsee
You can try
pandas.Series.str.catimport 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=paste0Or 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 loopI have used
itertoolsto mimic recyclingimport 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 respatsymight be relevant (not an experienced user myself.)
你可以试试
pandas.Series.str.catimport 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要不就
#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我曾经
itertools模仿回收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 respatsy可能是相关的（我自己不是有经验的用户。）
回答by Corey Levinson
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:
如果您只想将两个字符串列粘贴在一起，您可以简化@shouldsee 的答案，因为您不需要创建该函数。例如，就我而言：
df['newcol'] = df['id_part_one'].str.cat(df['id_part_two'], sep='_')
It might be required for both Series to be of dtype objectin order to this (I haven't verified).
为此，可能需要两个系列都为 dtype object（我尚未验证）。
