因此,我已这个pandas.Dataframepandas.apply期望输出形状(传递的值的形状为(x,),指数暗示(X,Y))
C1 C2 C3 C4 C5 Start End C8
A 1 - - - 1 4 -
A 2 - - - 6 10 -
A 3 - - - 11 14 -
A 4 - - - 15 19 -
其中 - 是对象,开始是初始坐标和结束是每个元素的最终坐标。
我定义了这个函数来计算表中所有间隔的联合,在这个例子中它应该总和为[1,19] - {5}(基本上是一个包含所有包含元素的numpy数组)。
def coverage(table):
#return a dataframe with the coverage of each individual peptide in a protein
interval = (table.apply(lambda row : range(int(row['Start']),int(row['End'])+1),axis=1))]
#if there is only one peptide, return the range between its start and end positions
if len(table) == 1: return asarray(range(int(table['Start']),int(table['End'])+1))
#if there are more, unite all the intervals
if len(table) > 1:
return reduce(union1d,(list(interval)))
因此,我将该函数迭代地应用于多个DataFrame(第一个是A,然后是B,C等)。问题是,对于一些表失败,并赠送此错误:
Traceback (most recent call last):
File "At_coverage.py", line 37, in <module>
covdir[prot] = coverage(data)
File "At_coverage.py", line 21, in coverage
interval = (table.apply(lambda row : range(int(row['Start']),int(row['End'])+1),axis=1))
File "/usr/lib/python2.7/dist-packages/pandas/core/frame.py", line 3312, in apply
return self._apply_standard(f, axis, reduce=reduce)
File "/usr/lib/python2.7/dist-packages/pandas/core/frame.py", line 3417, in _apply_standard
result = self._constructor(data=results, index=index)
File "/usr/lib/python2.7/dist-packages/pandas/core/frame.py", line 201, in __init__
mgr = self._init_dict(data, index, columns, dtype=dtype)
File "/usr/lib/python2.7/dist-packages/pandas/core/frame.py", line 323, in _init_dict
dtype=dtype)
File "/usr/lib/python2.7/dist-packages/pandas/core/frame.py", line 4473, in _arrays_to_mgr
return create_block_manager_from_arrays(arrays, arr_names, axes)
File "/usr/lib/python2.7/dist-packages/pandas/core/internals.py", line 3760, in create_block_manager_from_arrays
construction_error(len(arrays), arrays[0].shape[1:], axes, e)
File "/usr/lib/python2.7/dist-packages/pandas/core/internals.py", line 3732, in construction_error
passed,implied))
ValueError: Shape of passed values is (7,), indices imply (7, 8)
与它未能在下列数据框:
Protein Peptide \
11106 sp|Q75W54|EBM_ARATH GJDGFJK
11107 sp|Q75W54|EBM_ARATH GJDGFJK
11108 sp|Q75W54|EBM_ARATH JJDPHJVSTFFDDYKR
11109 sp|Q75W54|EBM_ARATH JJDPHJVSTFFDDYKR
11110 sp|Q75W54|EBM_ARATH JNGEPJFJR
11111 sp|Q75W54|EBM_ARATH JNGEPJFJR
11112 sp|Q75W54|EBM_ARATH JNGEPJFJR
Fraction Count \
11106 AT_indark_IEX_fraction_18a_20150422.uniprot-pr... 2
11107 AT_indark_IEX_fraction_21a_20150422.uniprot-pr... 2
11108 AT_indark_IEX_fraction_18a_20150422.uniprot-pr... 2
11109 AT_indark_IEX_fraction_19a_20150422.uniprot-pr... 1
11110 AT_indark_IEX_fraction_19a_20150422.uniprot-pr... 2
11111 AT_indark_IEX_fraction_22a_20150422.uniprot-pr... 2
11112 AT_indark_IEX_fraction_25a_20150422.uniprot-pr... 2
Sequence Start End Length
11106 MAEIGKTVLDFGWIAARSTEVDVNGVQLTTTNPPAISSESRWMEAA... 577 584 944
11107 MAEIGKTVLDFGWIAARSTEVDVNGVQLTTTNPPAISSESRWMEAA... 577 584 944
11108 MAEIGKTVLDFGWIAARSTEVDVNGVQLTTTNPPAISSESRWMEAA... 210 226 944
11109 MAEIGKTVLDFGWIAARSTEVDVNGVQLTTTNPPAISSESRWMEAA... 210 226 944
11110 MAEIGKTVLDFGWIAARSTEVDVNGVQLTTTNPPAISSESRWMEAA... 344 353 944
11111 MAEIGKTVLDFGWIAARSTEVDVNGVQLTTTNPPAISSESRWMEAA... 344 353 944
11112 MAEIGKTVLDFGWIAARSTEVDVNGVQLTTTNPPAISSESRWMEAA... 344 353 944
[7 rows x 8 columns]
为了使它工作,我更换了第三行与
interval = (table.apply(lambda row : range(int(row['Start']),int(row['End'])+4),axis=1)).apply(lambda row: row[:-3])
,我注意到它也适用于任何其他数量比+1(尽管有一些人在它的另一个数据框崩溃后的环路。
所以这个解决方案是多余的和愚蠢的。 MY HYPOTHESIS是这个特定数据框中的行数匹配一些奇怪的参数(比如列数或类似的东西),这使Pandas试图简化某些东西然后崩溃。
我做的,也适用于多次启动和结束程序的简化版本:
def multicov(row):
intervals = []
for i in range(len(row['Start'])):
#print data
intervals.append((range(int(row['Start'][i]),int(row['End'][i])+1)))
return reduce(union1d,intervals)
dir = {'Start':[[1,7],[14]],
'End':[[5,10],[18]]}
df = DataFrame(dir,columns=['Start','End'])
print df
print df.apply(multicov,axis=1)
在这种情况下,赠送了同样的错误
ValueError: Shape of passed values is (2,), indices imply (2, 2)
但有趣的是,如果我回到函数中的两个元素(以便它匹配2,2)表现良好。
return reduce(union1d,intervals),'foobar'
Start End
0 [1, 7] [5, 10]
1 [14] [18]
[2 rows x 2 columns]
0 ([1, 2, 3, 4, 7, 8, 9, 10], foobar)
1 ([14, 15, 16, 17, 18], foobar)
dtype: object
如果我指定输出作为一个列表,
return [reduce(union1d,intervals),'foobar']
它前面的列名的输出相匹配!
Start End
0 [1, 7] [5, 10]
1 [14] [18]
[2 rows x 2 columns]
Start End
0 [1, 2, 3, 4, 7, 8, 9] foobar
1 [14, 15, 16, 17] foobar
[2 rows x 2 columns]
所以我认为错误与熊猫试图迫使我以前的数据帧,并从输出一个与一些兼容性做的,但我很惊讶,对于大多数DataFrames它工作得很好!
为什么使用元组()不是Python的?我被告知永远不要在熊猫上迭代,这不是非常优化的。 – Nico
@Nico是的,矢量化的代码比循环要快得多。但是在这里,'apply'只是遍历行,不能被cython化或向量化。而且,创建未使用的数据帧还有一些开销,并且使用union1d进行多次调用,这会多次调用。 – ptrj