从元组列表创建熊猫数据框

Boosted_d16 发表于 Dev

Boosted_d16

我看过几个类似的主题，但是他们对我的新帖子并没有真正帮助我。

我想从元组列表中创建以下df：

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

这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

耶斯列尔

我认为您必须将一个值添加到列list，然后尝试list comprehension再set_index添加第一列，如果需要的话，请输入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，您可以使用解决方案评论的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