Python 如何从字符串列表中的每个字符串中删除最后一个字符
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How to delete the very last character from every string in a list of strings
提问by spartmar
I have the strings '80010', '80030', '80050' in a list, as in
我在列表中有字符串 '80010'、'80030'、'80050'，如
test = ['80010','80030','80050']
How can I delete the very last character (in this case the very last digit of each string which is a 0), so that I can end up with another list containing only the first four digits/characters from each string? So end up with something like
如何删除最后一个字符（在这种情况下，每个字符串的最后一个数字为 0），以便我可以得到另一个仅包含每个字符串前四位数字/字符的列表？所以最终得到类似的东西
newtest = ['8001', '8003', '8005']
I am very new to Python but I have tried with if-else statements, appending, using indexing [:-1], etc. but nothing seems to work unless I end up deleting all my other zeros. Thank you so much!
我对 Python 非常陌生，但我尝试过使用 if-else 语句、附加、使用索引 [:-1] 等，但除非我最终删除所有其他零，否则似乎没有任何效果。非常感谢！
采纳答案by Matthew
test = ["80010","80030","80050"]
newtest = [x[:-1] for x in test]
New test will contain the result ["8001","8003","8005"].
新测试将包含结果["8001","8003","8005"]。
[x[:-1] for x in test]creates a new list (using list comprehension) by looping over each item in testand putting a modified version into newtest. The x[:-1]means to take everything in the string value x up to but not including the last element.
[x[:-1] for x in test]创建一个新的列表（使用列表中理解通过遍历每个项目）test，并把修改后的版本为newtest。将x[:-1]字符串值 x 中的所有内容取到但不包括最后一个元素的方法。
回答by timgeb
You are not so far off. Using the slice notation [:-1] is the right approach. Just combine it with a list comprehension:
你并没有那么远。使用切片符号 [:-1] 是正确的方法。只需将它与列表理解结合起来：
>>> test = ['80010','80030','80050']
>>> [x[:-1] for x in test]
['8001', '8003', '8005']
somestring[:-1]gives you everything from the character at position 0 (inclusive) to the last character (exclusive).
somestring[:-1]为您提供从位置 0（包含）的字符到最后一个字符（不包含）的所有内容。
回答by Rockybilly
Just to show a slightly different solution than comprehension, Given that other answers already explained slicing, I just go through at the method.
只是为了展示一个与理解略有不同的解决方案，鉴于其他答案已经解释了切片，我只介绍了该方法。
With the map function.
带地图功能。
test = ['80010','80030','80050']
print map(lambda x: x[:-1],test)
# ['8001', '8003', '8005']
For more information about this solution, please read the brief explanation I did in another similar question.
有关此解决方案的更多信息，请阅读我在另一个类似问题中所做的简要说明。
回答by NRagot
In python @Matthew solution is perfect. But if indeed you are a beginer in coding in general, I must recommend this, less elegant for sure but the only way in many other scenario :
在python中@Matthew解决方案是完美的。但是，如果您确实是一般编码的初学者，我必须推荐这个，肯定不太优雅，但这是许多其他情况下的唯一方法：
#variables declaration
test = ['80010','80030','80050']
lenght = len(test) # for reading and writing sakes, len(A): lenght of A
newtest = [None] * lenght # newtest = [none, none, none], go look up empty array creation
strLen = 0 # temporary storage
#adding in newtest every element of test but spliced
for i in range(0, lenght): # for loop
str = test[i] # get n th element of test
strLen = len (str) # for reading sake, the lenght of string that will be spliced
newtest[i] = str[0:strLen - 1] # n th element of newtest is the spliced n th element from test
#show the results
print (newtest) # ['8001','8003','8005']
ps : this scripts, albeit not being the best, works in python ! Good luck to any programmer newcommer.
ps：这个脚本虽然不是最好的，但可以在 python 中运行！祝任何程序员新人好运。
