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将列表列表导入Pandas DataFrame

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问题：将列表列表导入Pandas DataFrame
回答 0
回答 1
回答 2

问题：将列表列表导入Pandas DataFrame

我正在将电子表格的内容读入熊猫。DataNitro具有一种返回矩形单元格选择作为列表列表的方法。所以

table = Cell("A1").table

给

table = [['Heading1', 'Heading2'], [1 , 2], [3, 4]]

headers = table.pop(0) # gives the headers as list and leaves data

我正忙着编写代码来翻译此内容，但我猜想它是如此简单，因此必须有方法可以做到这一点。无法在文档中找到它。任何可以简化此方法的指针？

点击查看英文原文

I am reading contents of a spreadsheet into pandas. DataNitro has a method that returns a rectangular selection of cells as a list of lists. So

table = Cell("A1").table

gives

table = [['Heading1', 'Heading2'], [1 , 2], [3, 4]]

headers = table.pop(0) # gives the headers as list and leaves data

I am busy writing code to translate this, but my guess is that it is such a simple use that there must be method to do this. Cant seem to find it in documentation. Any pointers to the method that would simplify this?

回答 0

pd.DataFrame直接调用构造函数：

df = pd.DataFrame(table, columns=headers)
df

   Heading1  Heading2
0         1         2
1         3         4
点击查看英文原文

Call the pd.DataFrame constructor directly:

df = pd.DataFrame(table, columns=headers)
df

   Heading1  Heading2
0         1         2
1         3         4

回答 1

使用上面EdChum解释的方法，列表中的值显示为行。要在DataFrame中将列表的值显示为列，只需使用transpose（）如下：

table = [[1 , 2], [3, 4]]
df = DataFrame(table)
df = df.transpose()
df.columns = ['Heading1', 'Heading2']

输出为：

      Heading1  Heading2
0         1        3
1         2        4
点击查看英文原文

With approach explained by EdChum above, the values in the list are shown as rows. To show the values of lists as columns in DataFrame instead, simply use transpose() as following:

table = [[1 , 2], [3, 4]]
df = DataFrame(table)
df = df.transpose()
df.columns = ['Heading1', 'Heading2']

The output then is:

      Heading1  Heading2
0         1        3
1         2        4

回答 2

即使没有pop列表，我们也可以set_index 

pd.DataFrame(table).T.set_index(0).T
Out[11]: 
0 Heading1 Heading2
1        1        2
2        3        4

更新资料 from_records

table = [['Heading1', 'Heading2'], [1 , 2], [3, 4]]

pd.DataFrame.from_records(table[1:],columns=table[0])
Out[58]: 
   Heading1  Heading2
0         1         2
1         3         4
点击查看英文原文

Even without pop the list we can do with set_index 

pd.DataFrame(table).T.set_index(0).T
Out[11]: 
0 Heading1 Heading2
1        1        2
2        3        4

Update from_records

table = [['Heading1', 'Heading2'], [1 , 2], [3, 4]]

pd.DataFrame.from_records(table[1:],columns=table[0])
Out[58]: 
   Heading1  Heading2
0         1         2
1         3         4