我有一个熊猫数据框架
>>> df
Out[126]:
score id
0 0.999989 654153
1 0.992971 941351
2 0.979518 701608
3 0.972667 564000
4 0.936928 999843并希望转换为可移植的(以便写入具有更好可读性的文本文件)
import prettytable as pt
x = pt.PrettyTable()
for col in list(df.columns):
x.add_column(col,df[col])然后在一个函数中,我使用
print(x.get_string())得到这个错误
File "<ipython-input-130-8db747160a67>", line 5, in <module>
verbose = True)
File "<ipython-input-129-4e27c067e0b5>", line 104, in lda_save_eval
print(x.get_string())
File "C:\Users\USER\Anaconda3\envs\tensorflow\lib\site-packages\prettytable.py", line 990, in get_string
self._compute_widths(formatted_rows, options)
File "C:\Users\USER\Anaconda3\envs\tensorflow\lib\site-packages\prettytable.py", line 894, in _compute_widths
widths = [_get_size(field)[0] for field in self._field_names]
File "C:\Users\USER\Anaconda3\envs\tensorflow\lib\site-packages\prettytable.py", line 894, in <listcomp>
widths = [_get_size(field)[0] for field in self._field_names]
File "C:\Users\USER\Anaconda3\envs\tensorflow\lib\site-packages\prettytable.py", line 77, in _get_size
lines = text.split("\n")
AttributeError: 'int' object has no attribute 'split'有什么线索吗?
发布于 2018-06-25 08:02:11
尝试print (x.get_string())而不是print(x.get_string)
为了改进这一点,为了使文本文件具有更好的可读性--您不需要循环使用tabulate --它给您提供了更多的灵活性。
尝尝这个,
from tabulate import tabulate
print (tabulate(df,df.columns,tablefmt='psql'))在tablefmt中,您可以提供许多选项来获得不同的样式。有关更多详细信息,请参阅此链接
https://stackoverflow.com/questions/51018046
复制相似问题