如果位置标签总是只有一个字符，简单的解决方案是这样的：

 >>> df.apply（lambda c：c.str [： -  2]）
 name1 name2 
 0 Marc Gasol Lebron James 
 1保罗加索尔凯文杜兰特
 2德怀特霍华德凯里欧文
 

系列的 str 属性可让您字符串操作，包括索引，所以这只是修剪每个值的最后两个字符。

关于你的问题关于 df.column ，这个问题比熊猫更普遍。这两件事情是不一样的：

 ＃works 
 obj.attr 

＃不工作
 attrName ='attr'
 obj.attrName 
 

你当您想访问其名称存储在变量中的属性时，不能使用点符号。通常，您可以使用 getattr 函数。然而，熊猫通过将名称指定为字符串（而不是源代码标识符）来提供访问列的括号符号。所以这两个是等价的：

  df.some_column 

 columnName =some_column
 df [columnName] 
 

在您的示例中，将引用更改为 df.column 到 df [column] 应该解决这个问题。但是，正如我在评论中提到的，您的代码也有其他问题。就解决手头的任务而言，我在回答开始时所显示的字符串索引方法要简单得多。

I have a DataFrame that I took from basketball-reference with player names. The code below is how I built the DataFrame. It has 5 columns of player names, but each name also has the player's position.

url = "http://www.basketball-reference.com/awards/all_league.html"
dframe_list = pd.io.html.read_html(url)
df = dframe_list[0]
df.drop(df.columns[[0,1,2]], inplace=True, axis=1)
column_names = ['name1', 'name2', 'name3', 'name4', 'name5']
df.columns = column_names
df = df[df.name1.notnull()]

I am trying to split off the position. To do so I had planned to make a DataFrame for each name column:

name1 = pd.DataFrame(df.name1.str.split().tolist()).ix[:,0:1]
name1[0] = name1[0] + " " + name1[1]
name1.drop(name1.columns[[1]], inplace=True, axis=1)

Since I have five columns I thought I would do this with a loop

column_names = ['name1', 'name2', 'name3', 'name4', 'name5']
for column in column_names:
    column = pd.DataFrame(df.column.str.split().tolist()).ix[:,0:1]
    column[0] = column[0] + " " + column[1]
    column.drop(column.columns[[1]], inplace=True, axis=1)
    column.columns = column

And then I'd join all these DataFrames back together.

df_NBA = [name1, name2, name3, name4, name5]
df_NBA = pd.concat(df_NBA, axis=1)

I'm new to python, so I'm sure I'm doing this in a pretty cumbersome fashion and would love suggestions as to how I might do this faster. But my main question is, when I run the code on individual columns it works fine, but if when I run the loop I get the error:

AttributeError: 'DataFrame' object has no attribute 'column'

It seems that the part of the loop df.column.str is causing some problem? I've fiddled around with the list, with bracketing column (I still don't understand why sometimes I bracket a DataFrame column and sometimes it's .column, but that's a bigger issue) and other random things.

When I try @BrenBarn's suggestion

df.apply(lambda c: c.str[:-2])

The following pops up in the Jupyter notebook:

SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame

See the caveats in the documentation:    http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
  if __name__ == '__main__':

Looking at the DataFrame, nothing has actually changed and if I understand the documentation correctly this method creates a copy of the DataFrame with the edits, but that this is a temporary copy that get's thrown out afterward so the actual DataFrame doesn't change.

If the position labels are always only one character, the simple solution is this:

>>> df.apply(lambda c: c.str[:-2])
           name1         name2
0     Marc Gasol  Lebron James
1      Pau Gasol  Kevin Durant
2  Dwight Howard  Kyrie Irving

The str attribute of a Series lets you do string operations, including indexing, so this just trims the last two characters off each value.

As for your question about df.column, this issue is more general than pandas. These two things are not the same:

# works
obj.attr

# doesn't work
attrName = 'attr'
obj.attrName

You can't use the dot notation when you want to access an attribute whose name is stored in a variable. In general, you can use the getattr function instead. However, pandas provides the bracket notation for accessing a column by specifying the name as a string (rather than a source-code identifier). So these two are equivalent:

df.some_column

columnName = "some_column"
df[columnName]

In your example, changing your reference to df.column to df[column] should resolve that issue. However, as I mentioned in a comment, your code has other problems too. As far as solving the task at hand, the string-indexing approach I showed at the beginning of my answer is much simpler.