Question

我正试图作一个扼杀,并将之附在一份清单中的每一条插图上,然后有一份新清单,其中附有完整的插图。例:

list1 = [ foo ,  fob ,  faz ,  funk ]
string =  bar 

*magic*

list2 = [ foobar ,  fobbar ,  fazbar ,  funkbar ]

我曾尝试过路,试图列出谅解,但这是垃圾。一如既往,任何帮助都受到高度赞赏。

Answer 1

这样做的最简单方式是罗列:

[s + mystring for s in mylist]

通知说,我避免使用诸如<代码>>><>>>/代码”等已建名称,因为这种影子或隐藏了所建名称,这并不好。

此外,如果你实际上不需要清单,但仅仅需要一个召集人,则发电机的表述可以提高效率(尽管在短名单上不可能如此):

(s + mystring for s in mylist)

这些都是非常有力、灵活和简明的。每名优秀学生都应学会学习,以学习他们。

Answer 2

my_list = [ foo ,  fob ,  faz ,  funk ]
string =  bar 
my_new_list = [x + string for x in my_list]
print my_new_list

这将印刷:

[ foobar ,  fobbar ,  fazbar ,  funkbar ]

Answer 3

似乎与我工作的适当工具一样。

my_list = [ foo ,  fob ,  faz ,  funk ]
string =  bar 
list2 = list(map(lambda orig_string: orig_string + string, my_list))

见,关于功能性方案拟订工具的一节<>,供更多实例使用map。

Answer 4

此处采用<代码>pandas的简单答案。

import pandas as pd
list1 = [ foo ,  fob ,  faz ,  funk ]
string =  bar 

list2 = (pd.Series(list1) + string).tolist()
list2
# [ foobar ,  fobbar ,  fazbar ,  funkbar ]

Answer 5

www.un.org/Depts/DGACM/index_spanish.htm 更新更多备选办法

以下是我所遵循的一些方法,我确信,可能有更多的方法。

<><><>>Method 1:>>>

list1 = [ foo ,  fob ,  faz ,  funk ]
list2 = [ls+"bar" for ls in list1] # using list comprehension
print(list2)

<><><>>Method2:>>>

list1 = [ foo ,  fob ,  faz ,  funk ]
list2 = list(map(lambda ls: ls+"bar", list1))
print(list2)

<><>><>Method 3:<>>>

list1 = [ foo ,  fob ,  faz ,  funk ]
addstring =  bar 
for index, value in enumerate(list1):
    list1[index] = addstring + value #this will prepend the string
    #list1[index] = value + addstring #this will append the string

<><><>Method 4:>>

list1 = [ foo ,  fob ,  faz ,  funk ]
addstring =  bar 
list2 = []
for value in list1:
    list2.append(str(value) + "bar")
print(list2)

<><><>Method 5:>>

list1 = [ foo ,  fob ,  faz ,  funk ]
list2 = list(map(  .join, zip(list1, ["bar"]*len(list1))))
print(list2)

避免使用关键词作为清单等变数,改称清单1

Answer 6

3. 试验如下:

[s + mystring for s in mylist]

似乎比明显使用像样的休闲设施要快35%:

i = 0
for s in mylist:
    mylist[i] = s+mystring
    i = i + 1

<>Experiment>

import random
import string
import time

mystring =  /test/ 

l = []
ref_list = []

for i in xrange( 10**6 ):
    ref_list.append(   .join(random.choice(string.ascii_lowercase) for i in range(10)) )

for numOfElements in [5, 10, 15 ]:

    l = ref_list*numOfElements
    print  Number of elements: , len(l)

    l1 = list( l )
    l2 = list( l )

    # Method A
    start_time = time.time()
    l2 = [s + mystring for s in l2]
    stop_time = time.time()
    dt1 = stop_time - start_time
    del l2
    #~ print "Method A: %s seconds" % (dt1)

    # Method B
    start_time = time.time()
    i = 0
    for s in l1:
        l1[i] = s+mystring
        i = i + 1
    stop_time = time.time()
    dt0 = stop_time - start_time
    del l1
    del l
    #~ print "Method B: %s seconds" % (dt0)

    print  Method A is %.1f%% faster than Method B  % ((1 - dt1/dt0)*100)

<>Results

Number of elements: 5000000
Method A is 38.4% faster than Method B
Number of elements: 10000000
Method A is 33.8% faster than Method B
Number of elements: 15000000
Method A is 35.5% faster than Method B

Answer 7

缩略语:

>>> list(map( {}bar .format,  [ foo ,  fob ,  faz ,  funk ]))
[ foobar ,  fobbar ,  fazbar ,  funkbar ]

Thus, there is no loop variable.
It works for Python 2 and 3. (In Python 3 one can write [*map(...)], and in Python 2 just map(...).

如果赞成模块表达

>>> list(map( %sbar .__mod__,  [ foo ,  fob ,  faz ,  funk ]))
[ foobar ,  fobbar ,  fazbar ,  funkbar ]

<<>prepend/strong>, 可查阅__add__<>>>。方法


>>> list(map( bar .__add__,  [ foo ,  fob ,  faz ,  funk ]))
[ barfoo ,  barfob ,  barfaz ,  barfunk ]

Answer 8

将范围扩大至“将示意图清单列入示意图清单”:

    import numpy as np
    lst1 = [ a , b , c , d , e ]
    lst2 = [ 1 , 2 , 3 , 4 , 5 ]

    at = np.full(fill_value= @ ,shape=len(lst1),dtype=object) #optional third list
    result = np.array(lst1,dtype=object)+at+np.array(lst2,dtype=object)

<>Result>:

array([ a@1 ,  b@2 ,  c@3 ,  d@4 ,  e@5 ], dtype=object)

d 粉碎机可进一步改装

Answer 9

new_list = [word_in_list + end_string for word_in_list in old_list]

使用“名单”等贵方变量名称是不好的,因为清单将超出建筑群。

Answer 10

you can use lambda inside map in python. wrote a gray codes generator. https://github.com/rdm750/rdm750.github.io/blob/master/python/gray_code_generator.py # your code goes here the n-1 bit code, with 0 prepended to each word, followed by the n-1 bit code in reverse order, with 1 prepended to each word.

    def graycode(n):
        if n==1:
            return [ 0 , 1 ]
        else:
            nbit=map(lambda x: 0 +x,graycode(n-1))+map(lambda x: 1 +x,graycode(n-1)[::-1])
            return nbit

    for i in xrange(1,7):
        print map(int,graycode(i))

Answer 11

由于使用窒息法的3.6是最佳做法(而不是<代码>format或与+保持一致)。见PEP498。

list1 = [ foo ,  fob ,  faz ,  funk ]
mystring =  bar 

list2 = [f"{s}{mystring}" for s in list1]

Answer 12


list2 = [ %sbar  % (x,) for x in list]

并且不使用<代码>>list作为名称;它为所建类型投影。

Answer 13

案情陈述

list = [ foo ,  fob ,  faz ,  funk ]
string =  bar 
for i in range(len(list)):
    list[i] += string
print(list)

友情链接