如何将两个不同的元组列表取消列出到同一结构并附加到文件对象?

使用 zip

test_tup = [(111, 222, 333), ('aaa', 'bbb', 'ccc'), ('a1', 'b2', 'c3')]
print(list(zip(*test_tup)))

输出:

[(111, 'aaa', 'a1'), (222, 'bbb', 'b2'), (333, 'ccc', 'c3')]

把它们分开写在一起 ;

print('; '.join(' '.join(str(x) for x in i) for i in headers))
print('\n'.join('; '.join(str(x) for x in i) for i in res))

id; attr_1; attr_2
111; aaa; a1
222; bbb; b2
333; ccc; c3

如果要将所需输出写入csv文件,请尝试以下操作:

import pandas as pd
a = [('id',), ('attr_1',), ('attr_2',), ...]
b = [(111, 222, 333), ('aaa', 'bbb', 'ccc'), ('a1', 'b2', 'c3'), ...]
df = pd.DataFrame({a[i][0] : b[i] for i in range(len(a))}) # length of a and each item inside b must be same
df.to_csv(file_name, sep=';')

你可以这样做,

>>> header
[('id',), ('attr_1',), ('attr_2',)]
>>> x
[(111, 222, 333), ('aaa', 'bbb', 'ccc'), ('a1', 'b2', 'c3')]
>>> ';'.join(x[0] for x in header)
'id;attr_1;attr_2'
>>> print('\n'.join(';'.join(str(_) for _ in y) for y in zip(*x)))
111;aaa;a1
222;bbb;b2
333;ccc;c3

你不能 .join int s、 你应该把它们转换成 str 首先,然后使用 zip :

values = [(111, 222, 333), ('aaa', 'bbb', 'ccc'), ('a1', 'b2', 'c3')]
values = [tuple(str(j) for j in i) for i in values]
out = ['; '.join(i) for i in zip(*values)]
print(out) #['111; aaa; a1', '222; bbb; b2', '333; ccc; c3']

选择性地添加 out = '\n'.join(out) 一行接一行 out = ['; '.join(i) for i in zip(*values)] str

您也可以在第二个列表上执行join,但需要在列表的每个元素上执行join。F.ex按清单Compherence:

value_list = [ ";".join(i) for i in value]