如果不为空，则评估每个单元格和 return 列标题 pandas df

Question

我有 pandas.df 233 行 * 234 列，我需要评估每个单元格和 return 对应的列 header 如果不是 nan，到目前为止我写了以下内容：

#First get a list of all column names (except column 0):

col_list=[]

for column in df.columns[1:]:
    col_list.append(column)

#Then I try to iterate through every cell and evaluate for Null
#Also a counter is initiated to take the next col_name from col_list
#when count reach 233

for index, row in df.iterrows():
    count = 0
    for x in row[1:]:
        count = count+1
        for col_name in col_list:
            if count >= 233: break
            elif str(x) != 'nan':
                print col_name

代码并没有完全做到这一点，我需要更改什么才能让代码在 233 行后中断并转到下一个 col_name？

Example:

    Col_1   Col_2    Col_3
1    nan     13       nan
2    10      nan      nan
3    nan      2        5
4    nan     nan       4

output:      
1   Col_2
2   Col_1
3   Col_2
4   Col_3
5   Col_3

Answer 1

您可以使用 dropna :

df.dropna(axis=1).columns

轴：{0 或‘索引’，1 或‘列’}

如何：{‘任何’，‘全部’}

基本上是用dropna去掉null，axis=1是去掉列，how="any"去掉是至少有一个列是null，.columns得到剩下的header.

Answer 2

如果第一列是 index stack - it remove all NaNs and then get values from second level of Multiindex by reset_index and selecting or by Series constructor with Index.get_level_values:

我想你需要

s = df.stack().reset_index()['level_1'].rename('a')
print (s)
0    Col_2
1    Col_1
2    Col_2
3    Col_3
4    Col_3
Name: a, dtype: object

或者：

s = pd.Series(df.stack().index.get_level_values(1))
print (s)
0    Col_2
1    Col_1
2    Col_2
3    Col_3
4    Col_3
dtype: object

如果需要输出为list:

L = df.stack().index.get_level_values(1).tolist()
print (L)
['Col_2', 'Col_1', 'Col_2', 'Col_3', 'Col_3']

详情:

print (df.stack())
1  Col_2    13.0
2  Col_1    10.0
3  Col_2     2.0
   Col_3     5.0
4  Col_3     4.0
dtype: float64

Answer 3

我会使用 jezrael 的堆栈解决方案。

但是，如果您对 Numpy 方式感兴趣，这通常更快。

In [4889]: np.tile(df.columns, df.shape[0])[~np.isnan(df.values.ravel())]
Out[4889]: array(['Col_2', 'Col_1', 'Col_2', 'Col_3', 'Col_3'], dtype=object)

时间

In [4913]: df.shape
Out[4913]: (100, 3)

In [4914]: %timeit np.tile(df.columns, df.shape[0])[~np.isnan(df.values.ravel())]
10000 loops, best of 3: 35.8 µs per loop

In [4915]: %timeit df.stack().index.get_level_values(1)
1000 loops, best of 3: 335 µs per loop

In [4905]: df.shape
Out[4905]: (100000, 3)

In [4907]: %timeit np.tile(df.columns, df.shape[0])[~np.isnan(df.values.ravel())]
100 loops, best of 3: 5.98 ms per loop

In [4908]: %timeit df.stack().index.get_level_values(1)
100 loops, best of 3: 11.7 ms per loop

根据您的需要（可读性、速度、可维护性等）进行选择

如果不为空，则评估每个单元格和 return 列标题 pandas df

evaluate every cell and return column head if not null pandas df

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nested-loops

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