问题:在Python中将相同的字符串追加到字符串列表
我正在尝试采用一个字符串,并将其附加到列表中包含的每个字符串中,然后使用完成的字符串创建一个新列表。例:
list = ['foo', 'fob', 'faz', 'funk']
string = 'bar'
*magic*
list2 = ['foobar', 'fobbar', 'fazbar', 'funkbar']
我尝试了循环,并尝试了列表理解,但这是垃圾。一如既往的任何帮助,不胜感激。
I am trying to take one string, and append it to every string contained in a list, and then have a new list with the completed strings. Example:
list1 = ['foo', 'fob', 'faz', 'funk']
string = 'bar'
*magic*
list2 = ['foobar', 'fobbar', 'fazbar', 'funkbar']
I tried for loops, and an attempt at list comprehension, but it was garbage. As always, any help, much appreciated.
回答 0
最简单的方法是使用列表理解:
[s + mystring for s in mylist]
请注意,我避免使用内置名称,list
因为那样会掩盖或隐藏内置名称,这非常不好。
另外,如果您实际上不需要列表,而只需要一个迭代器,则生成器表达式可以更有效(尽管在短列表中这并不重要):
(s + mystring for s in mylist)
这些功能非常强大,灵活且简洁。每个好的python程序员都应该学会使用它们。
The simplest way to do this is with a list comprehension:
[s + mystring for s in mylist]
Notice that I avoided using builtin names like list
because that shadows or hides the builtin names, which is very much not good.
Also, if you do not actually need a list, but just need an iterator, a generator expression can be more efficient (although it does not likely matter on short lists):
(s + mystring for s in mylist)
These are very powerful, flexible, and concise. Every good python programmer should learn to wield them.
回答 1
my_list = ['foo', 'fob', 'faz', 'funk']
string = 'bar'
my_new_list = [x + string for x in my_list]
print my_new_list
这将打印:
['foobar', 'fobbar', 'fazbar', 'funkbar']
my_list = ['foo', 'fob', 'faz', 'funk']
string = 'bar'
my_new_list = [x + string for x in my_list]
print my_new_list
This will print:
['foobar', 'fobbar', 'fazbar', 'funkbar']
回答 2
map
对我来说,这似乎是工作的正确工具。
my_list = ['foo', 'fob', 'faz', 'funk']
string = 'bar'
list2 = list(map(lambda orig_string: orig_string + string, my_list))
有关的更多示例,请参见本节的函数编程工具map
。
map
seems like the right tool for the job to me.
my_list = ['foo', 'fob', 'faz', 'funk']
string = 'bar'
list2 = list(map(lambda orig_string: orig_string + string, my_list))
See this section on functional programming tools for more examples of map
.
回答 3
以pythonic方式运行以下实验:
[s + mystring for s in mylist]
似乎比明显的for循环使用快约35%:
i = 0
for s in mylist:
mylist[i] = s+mystring
i = i + 1
实验
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)
结果
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
Running the following experiment the pythonic way:
[s + mystring for s in mylist]
seems to be ~35% faster than the obvious use of a for loop like this:
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
回答 4
扩展到“将字符串列表追加到字符串列表”:
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)
结果:
array(['a@1', 'b@2', 'c@3', 'd@4', 'e@5'], dtype=object)
dtype odject可以进一步转换为str
Extending a bit to “Appending a list of strings to a list of strings”:
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)
dtype odject may be further converted str
回答 5
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))
回答 6
更新更多选项
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
避免将关键字用作“列表”之类的变量,而应将“列表”重命名为“ list1”
Updating with more options
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
Avoid using keywords as variables like ‘list’, renamed ‘list’ as ‘list1’ instead
回答 7
这是使用的简单答案pandas
。
import pandas as pd
list1 = ['foo', 'fob', 'faz', 'funk']
string = 'bar'
list2 = (pd.Series(list1) + string).tolist()
list2
# ['foobar', 'fobbar', 'fazbar', 'funkbar']
Here is a simple answer using pandas
.
import pandas as pd
list1 = ['foo', 'fob', 'faz', 'funk']
string = 'bar'
list2 = (pd.Series(list1) + string).tolist()
list2
# ['foobar', 'fobbar', 'fazbar', 'funkbar']
回答 8
list2 = ['%sbar' % (x,) for x in list]
并且不要使用list
名字;它隐藏了内置类型。
list2 = ['%sbar' % (x,) for x in list]
And don’t use list
as a name; it shadows the built-in type.
回答 9
new_list = [word_in_list + end_string for word_in_list in old_list]
为变量名使用“ list”之类的名称是不好的,因为它将覆盖/覆盖内置函数。
new_list = [word_in_list + end_string for word_in_list in old_list]
Using names such as “list” for your variable names is bad since it will overwrite/override the builtins.
回答 10
以防万一
list = ['foo', 'fob', 'faz', 'funk']
string = 'bar'
for i in range(len(list)):
list[i] += string
print(list)
Just in case
list = ['foo', 'fob', 'faz', 'funk']
string = 'bar'
for i in range(len(list)):
list[i] += string
print(list)