我是Python的新手,我只是想弄明白一些事情。
我有一堆txt文件要打开,其中包含数千个单词,全部由空格分隔,我相信我已经在下面的代码中介绍了这些空格。但我只是想知道如何将这些行分解成用户输入的3或一些int块?
来自['perfect', 'dealing', 'crave', 'stirring', 'expedition', 'alexandria', 'vanish', 'dealing', 'crave']
转到[['perfect', 'dealing', 'crave'], ['stirring', 'expedition', 'alexandria'],['vanish', 'dealing', 'crave']]
或者甚至返回到列表形成
a
b
c提亚
word_list=[]
filename = "filename.txt"
with open(filename,"r") as file_object:
for line in file_object:
word_list=line.split()
print(word_list)
return(word_list)发布于 2021-05-25 13:16:10
您可以使用for i in range(0, len(word_list), 3)遍历间隔为3的列表,并使用索引对原始列表进行切片
new_list = [word_list[i:i+3] for i in range(0, len(word_list), 3)]
# [['perfect', 'dealing', 'crave'], ['stirring', 'expedition', 'alexandria'], ['vanish', 'dealing', 'crave']]发布于 2021-05-25 13:28:23
你没有更新word_list,而是重写了它的值,它应该是这样的。
word_list=[]
filename = "filename.txt"
with open(filename,"r") as file_object:
for line in file_object:
word_list.append(line.split()) # notice this
print(word_list)
return(word_list)
sub_list_size = 3; # take from user input or as needed
output = [word_list[i: i+sub_list_size] for i in range(0, len(word_list), sub_list_size)]
print(output)发布于 2021-05-25 13:37:03
使用itertools配方中的grouper结构:
def grouper(iterable, n, fillvalue=None):
"Collect data into fixed-length chunks or blocks"
# grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
args = [iter(iterable)] * n
return zip_longest(*args, fillvalue=fillvalue)然后你的代码是:
word_list = []
n = 3
filename = "filename.txt"
with open(filename, "r") as file_object:
for line in file_object:
word_list.append(grouper(line.split(), n))
return word_listhttps://stackoverflow.com/questions/67681945
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