pandas - 从元组列表创建 Pandas 数据框

Question

我看过几个类似的帖子，但它们并没有真正帮助我，因此发布了新帖子。

我想从元组列表中创建下面的 df:

Values         Total  extra
label                      
Pictionary  0.000000     12
Chess       4.609929     12
Cluedo      8.421986     12

以下是实现它的所有组件:

columns = ['Total, 'extra']

tups = [(u'Pictionary', 0.0, 12)
        (u'Chess', 4.6099290780141837, 12)
        (u'Cluedo', 8.4219858156028362, 12)]

我失败的尝试:

pd.DataFrame(tups, columns=columns)

错误信息:

AssertionError: 2 columns passed, passed data had 3 columns

Answer 1

我认为您必须为列添加一个值 list然后尝试 list comprehension然后 set_index 第一列，如果需要第一列为 index :

import pandas as pd

columns = ['label', 'Total', 'extra']

tups = [(u'Pictionary', 0.0, 12),
        (u'Chess', 4.6099290780141837, 12),
        (u'Cluedo', 8.4219858156028362, 12)]

df = pd.DataFrame([x for x in tups], columns=columns)

print df
        label     Total  extra
0  Pictionary  0.000000     12
1       Chess  4.609929     12
2      Cluedo  8.421986     12

df = df.set_index('label')
#if you need set column name
df.columns.name = 'Values'

print df
Values         Total  extra
label                      
Pictionary  0.000000     12
Chess       4.609929     12
Cluedo      8.421986     12

或者您可以通过 comment 使用解决方案的 Colonel Beauvel :

import pandas as pd

columns = ['Total', 'extra']

tups = [(u'Pictionary', 0.0, 12),
        (u'Chess', 4.6099290780141837, 12),
        (u'Cluedo', 8.4219858156028362, 12)]

df = pd.DataFrame(tups, columns=['label']+columns)
print df
        label     Total  extra
0  Pictionary  0.000000     12
1       Chess  4.609929     12
2      Cluedo  8.421986     12

df = df.set_index('label')
df.columns.name = 'Values'
print df
Values         Total  extra
label                      
Pictionary  0.000000     12
Chess       4.609929     12
Cluedo      8.421986     12