在 Pandas 应用函数中获取行的索引 [英] getting the index of a row in a pandas apply function

问题描述

我正在尝试访问 Pandas 中应用于整个 DataFrame 的函数中的行索引.我有这样的事情:

I am trying to access the index of a row in a function applied across an entire DataFrame in Pandas. I have something like this:

df = pandas.DataFrame([[1,2,3],[4,5,6]], columns=['a','b','c'])
>>> df
   a  b  c
0  1  2  3
1  4  5  6

我将定义一个函数来访问给定行的元素

and I'll define a function that access elements with a given row

def rowFunc(row):
    return row['a'] + row['b'] * row['c']

我可以这样应用它:

df['d'] = df.apply(rowFunc, axis=1)
>>> df
   a  b  c   d
0  1  2  3   7
1  4  5  6  34

太棒了！现在如果我想将索引合并到我的函数中怎么办?在添加 d 之前，此 DataFrame 中任何给定行的索引将是 Index([u'a', u'b', u'c', u'd'], dtype='object')，但我想要 0 和 1.所以我不能只访问 row.index.

Awesome! Now what if I want to incorporate the index into my function? The index of any given row in this DataFrame before adding d would be Index([u'a', u'b', u'c', u'd'], dtype='object'), but I want the 0 and 1. So I can't just access row.index.

我知道我可以在存储索引的表中创建一个临时列，但我想知道它是否存储在某个行对象中.

I know I could create a temporary column in the table where I store the index, but I'm wondering if it is stored in the row object somewhere.

推荐答案

在这种情况下，要访问索引，请访问 name 属性:

To access the index in this case you access the name attribute:

In [182]:

df = pd.DataFrame([[1,2,3],[4,5,6]], columns=['a','b','c'])
def rowFunc(row):
    return row['a'] + row['b'] * row['c']

def rowIndex(row):
    return row.name
df['d'] = df.apply(rowFunc, axis=1)
df['rowIndex'] = df.apply(rowIndex, axis=1)
df
Out[182]:
   a  b  c   d  rowIndex
0  1  2  3   7         0
1  4  5  6  34         1

请注意，如果这确实是您要尝试执行的操作，那么以下操作会有效且速度更快:

Note that if this is really what you are trying to do that the following works and is much faster:

In [198]:

df['d'] = df['a'] + df['b'] * df['c']
df
Out[198]:
   a  b  c   d
0  1  2  3   7
1  4  5  6  34

In [199]:

%timeit df['a'] + df['b'] * df['c']
%timeit df.apply(rowIndex, axis=1)
10000 loops, best of 3: 163 µs per loop
1000 loops, best of 3: 286 µs per loop

编辑

3 年多以后再看这个问题，你可以这样做:

Looking at this question 3+ years later, you could just do:

In[15]:
df['d'],df['rowIndex'] = df['a'] + df['b'] * df['c'], df.index
df

Out[15]: 
   a  b  c   d  rowIndex
0  1  2  3   7         0
1  4  5  6  34         1

但假设它不是那么简单，无论您的 rowFunc 真正在做什么，您都应该考虑使用向量化函数，然后将它们用于 df 索引:

but assuming it isn't as trivial as this, whatever your rowFunc is really doing, you should look to use the vectorised functions, and then use them against the df index:

In[16]:
df['newCol'] = df['a'] + df['b'] + df['c'] + df.index
df

Out[16]: 
   a  b  c   d  rowIndex  newCol
0  1  2  3   7         0       6
1  4  5  6  34         1      16

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