我如何写一个列表文件?writelines()不插入换行符,所以我需要这样做:
f.writelines([f"{line}\n" for line in lines])
我如何写一个列表文件?writelines()不插入换行符,所以我需要这样做:
f.writelines([f"{line}\n" for line in lines])
当前回答
序列化列表到文本文件与逗号分隔值
mylist = dir()
with open('filename.txt','w') as f:
f.write( ','.join( mylist ) )
其他回答
因为我很懒....
import json
a = [1,2,3]
with open('test.txt', 'w') as f:
f.write(json.dumps(a))
#Now read the file back into a Python list object
with open('test.txt', 'r') as f:
a = json.loads(f.read())
简单:
with open("text.txt", 'w') as file:
file.write('\n'.join(yourList))
使用循环:
with open('your_file.txt', 'w') as f:
for line in lines:
f.write(f"{line}\n")
对于Python <3.6:
with open('your_file.txt', 'w') as f:
for line in lines:
f.write("%s\n" % line)
对于Python 2,还可以使用:
with open('your_file.txt', 'w') as f:
for line in lines:
print >> f, line
如果您热衷于单个函数调用,至少要删除方括号[],以便每次生成一个要打印的字符串(genexp而不是listcomp)——没有理由占用物化整个字符串列表所需的所有内存。
设avg为列表,则:
In [29]: a = n.array((avg))
In [31]: a.tofile('avgpoints.dat',sep='\n',dtype = '%f')
您可以根据自己的需求使用%e或%s。
更简单的是:
with open("outfile", "w") as outfile:
outfile.write("\n".join(itemlist))
要确保项目列表中的所有项目都是字符串,请使用生成器表达式:
with open("outfile", "w") as outfile:
outfile.write("\n".join(str(item) for item in itemlist))
记住,项目列表占用内存,所以要注意内存消耗。