如何将文本板中的所有行显示到列表中。Python3错误
我想要实现的是尝试用文本打开一个文本文件。文本的格式如下所示:如何将文本板中的所有行显示到列表中。Python3错误,python,list,python-3.x,Python,List,Python 3.x,我想要实现的是尝试用文本打开一个文本文件。文本的格式如下所示: Bear Car Plant 等等。现在我有了代码: try: with open(info) as information: for line in information.readlines(): line = line.split() except IOError as error: print("Failed to open. Try again.") 仅打印列表中
Bear
Car
Plant
等等。现在我有了代码:
try:
with open(info) as information:
for line in information.readlines():
line = line.split()
except IOError as error:
print("Failed to open. Try again.")
仅打印列表中的最后一行。我要做的是打印出列表中的所有单词
因此,对于上面的示例,当我打印(行)时,它打印['Plant']
,但我想要['Bear'、'Car'、'Plant']
谁能把我引向正确的方向 您不需要拆分,您需要使用剥离(删除换行符),然后将结果添加到列表中:
lines = []
with open(info) as information:
for line in information:
lines.append(line.strip())
请注意,根本不需要调用file.readlines()
;只需迭代文件就足够了
您可以通过读取整个文件并使用以下命令一次性完成此操作:
或者您也可以在列表中使用迭代:
with open(info) as information:
lines = [line.strip() for line in information]
readlines
将文件作为列表读取
try:
with open(info) as information:
lines = information.readlines()
except IOError as error:
print("Failed to open. Try again.")
您还可以使用列表
:
try:
with open(info) as information:
lines = list(information)
except IOError as error:
print("Failed to open. Try again.")
try:
with open(info) as information:
lines = list(information)
except IOError as error:
print("Failed to open. Try again.")