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ValueError:传递值的形状为(1,6),索引暗示(6,6)

valueerror flask pandas python

53

Sbk3824 · 技术社区 · 8 年前

我将一个列表从flask函数传递到另一个函数,并得到这个值错误。

发送端我的代码:

@app.route('/process', methods=['POST'])
def process():
    name = request.form['name']
    comment = request.form['comment']
    wickets = request.form['wickets']
    ga = request.form['ga']
    ppballs = request.form['ppballs']
    overs = request.form['overs']

    score = [name,comment,wickets,ga,ppballs,overs]
    results = []
    results = eval_score(score)
    print results

接收端:

def ml_model(data):
    col = pd.DataFrame(data,columns=['runs','balls', 'wickets', 'ground_average', 'pp_balls_left', 'total_overs'])
    predicted = predictor(col)

 ...
 line 1598, in dispatch_request
 return self.view_functions[rule.endpoint](**req.view_args)

 File "/Users/sbk/guestbook/guestbook.py", line 26, in process
 results = eval_score(score)

 File "/Users/sbk/guestbook/eval_score.py", line 6, in eval_score
 col = pd.DataFrame(data,columns=['runs','balls', 'wickets',  'ground_average', 'pp_balls_left', 'total_overs'])

 File "/Users/sbk/anaconda2/lib/python2.7/site-  packages/pandas/core/frame.py", line 385, in __init__
 copy=copy)

 File "/Users/sbk/anaconda2/lib/python2.7/site-packages/pandas/core/frame.py", line 533, in _init_ndarray
 return create_block_manager_from_blocks([values], [columns, index])

 File "/Users/sbk/anaconda2/lib/python2.7/site-packages/pandas/core/internals.py", line 4631, in  create_block_manager_from_blocks
 construction_error(tot_items, blocks[0].shape[1:], axes, e)

 File "/Users/sbk/anaconda2/lib/python2.7/site-packages/pandas/core/internals.py", line 4608, in construction_error
 Open an interactive python shell in this framepassed, implied))

请告诉我哪里出了问题。

4 回复 | 直到 8 年前

1

112

rafaelc 8 年前

只需更改

col = pd.DataFrame(data, columns=['runs','balls', 'wickets', 'ground_average', 'pp_balls_left', 'total_overs'])

对于

col = pd.DataFrame([data], columns=['runs','balls', 'wickets', 'ground_average', 'pp_balls_left', 'total_overs'])

你想要的 [data] 对于 pandas 要理解它们是行。

简单说明:

a = [1, 2, 3]
>>> pd.DataFrame(a)
   0
0  1
1  2
2  3

>>> pd.DataFrame([a])
   0  1  2
0  1  2  3

2

4

DevLoverUmar 5 年前

我在邮件中遇到了类似的错误

传递值的形状为(681783),索引在数据帧中暗示(6868)

根据我的猜测,我输入了数据数组的转置,这就解决了问题

更改自

Features_Dataframe = pd.DataFrame(data=Features, columns=Feature_Labels)  # here Features ndarray is 68*1783

到

Features_Dataframe = pd.DataFrame(data=Features.transpose(), columns=Feature_Labels)  # Now Features array became 1783*68 i.e., 1783 rows and 68 columns

3

Candice Withrow 6 年前

我在从回归系数(regressor.coeff\)生成数据帧时遇到了类似的问题,括号中给出了另一个要求二维输入的错误。如果出现此错误,请尝试在输入数组中附加[0],以便将值取出。例如: 数据[0]

4

0

Anabel Montero Posada 4 年前

pd。DataFrame(regr.coef\u,df\u coeficients.columns,columns=['coefficient'])

对此:

pd。DataFrame(regr.coef\u[0],df\u coeficients.columns,columns=['coefficient'])