问题：从python pandas中的列名获取列索引
在R中，当您需要根据列名检索列索引时，可以执行此操作
idx <- which(names(my_data)==my_colum_name)
有没有办法对熊猫数据框做同样的事情？
In R when you need to retrieve a column index based on the name of the column you could do
idx <- which(names(my_data)==my_colum_name)
Is there a way to do the same with pandas dataframes?
回答 0
当然可以使用.get_loc()：
In [45]: df = DataFrame({"pear": [1,2,3], "apple": [2,3,4], "orange": [3,4,5]})
In [46]: df.columns
Out[46]: Index([apple, orange, pear], dtype=object)
In [47]: df.columns.get_loc("pear")
Out[47]: 2
虽然老实说，我自己通常不需要这个。通常，通过名称进行访问可以实现我想要的功能（df["pear"]，，df[["apple", "orange"]]或者也许df.columns.isin(["orange", "pear"])），尽管我可以肯定地看到一些您需要索引号的情况。
Sure, you can use .get_loc():
In [45]: df = DataFrame({"pear": [1,2,3], "apple": [2,3,4], "orange": [3,4,5]})
In [46]: df.columns
Out[46]: Index([apple, orange, pear], dtype=object)
In [47]: df.columns.get_loc("pear")
Out[47]: 2
although to be honest I don’t often need this myself. Usually access by name does what I want it to (df["pear"], df[["apple", "orange"]], or maybe df.columns.isin(["orange", "pear"])), although I can definitely see cases where you’d want the index number.
回答 1
这是通过列表理解的解决方案。cols是要获取索引的列的列表：
[df.columns.get_loc(c) for c in cols if c in df]
Here is a solution through list comprehension. cols is the list of columns to get index for:
[df.columns.get_loc(c) for c in cols if c in df]
回答 2
DSM的解决方案有效，但是如果您想要直接等效的解决方案，则which可以这样做(df.columns == name).nonzero()
DSM’s solution works, but if you wanted a direct equivalent to which you could do (df.columns == name).nonzero()
回答 3
当您要查找多个列匹配项时，可以使用使用searchsorted方法的矢量化解决方案。因此，df作为query_cols要搜索的数据框和列名，实现将是-
def column_index(df, query_cols):
cols = df.columns.values
sidx = np.argsort(cols)
return sidx[np.searchsorted(cols,query_cols,sorter=sidx)]
样品运行-
In [162]: df
Out[162]:
apple banana pear orange peach
0 8 3 4 4 2
1 4 4 3 0 1
2 1 2 6 8 1
In [163]: column_index(df, ['peach', 'banana', 'apple'])
Out[163]: array([4, 1, 0])
When you might be looking to find multiple column matches, a vectorized solution using searchsorted method could be used. Thus, with df as the dataframe and query_cols as the column names to be searched for, an implementation would be –
def column_index(df, query_cols):
cols = df.columns.values
sidx = np.argsort(cols)
return sidx[np.searchsorted(cols,query_cols,sorter=sidx)]
Sample run –
In [162]: df
Out[162]:
apple banana pear orange peach
0 8 3 4 4 2
1 4 4 3 0 1
2 1 2 6 8 1
In [163]: column_index(df, ['peach', 'banana', 'apple'])
Out[163]: array([4, 1, 0])
回答 4
如果您希望从列位置获得列名（OP问题的另一种方法），则可以使用：
>>> df.columns.get_values()[location]
使用@DSM示例：
>>> df = DataFrame({"pear": [1,2,3], "apple": [2,3,4], "orange": [3,4,5]})
>>> df.columns
Index(['apple', 'orange', 'pear'], dtype='object')
>>> df.columns.get_values()[1]
'orange'
其他方法：
df.iloc[:,1].name
df.columns[location] #(thanks to @roobie-nuby for pointing that out in comments.)
In case you want the column name from the column location (the other way around to the OP question), you can use:
>>> df.columns.get_values()[location]
Using @DSM Example:
>>> df = DataFrame({"pear": [1,2,3], "apple": [2,3,4], "orange": [3,4,5]})
>>> df.columns
Index(['apple', 'orange', 'pear'], dtype='object')
>>> df.columns.get_values()[1]
'orange'
Other ways:
df.iloc[:,1].name
df.columns[location] #(thanks to @roobie-nuby for pointing that out in comments.)
回答 5
这个怎么样：
df = DataFrame({"pear": [1,2,3], "apple": [2,3,4], "orange": [3,4,5]})
out = np.argwhere(df.columns.isin(['apple', 'orange'])).ravel()
print(out)
[1 2]
how about this:
df = DataFrame({"pear": [1,2,3], "apple": [2,3,4], "orange": [3,4,5]})
out = np.argwhere(df.columns.isin(['apple', 'orange'])).ravel()
print(out)
[1 2]
