Question

我有一个像这样的数据框（df1）：

id  detail
78  [{}{}{}{}{}]
120 [{}{}{}{}{}]
110 [{}{}{}{}{}]
109 [{}{}{}{}{}]
109 [{}{}{}{}{}]
79  [{}{}{}{}{}]

详细信息列包含词典列表，每个词典如下所示：

{'y1': 549, 'score': 1, 'x2': 630, 'frame': 1054, 'y2': 564, 'x1': 602, 'visibility': 0.0, 'class': 5}

我需要将此信息提取为以下格式的CSV文件：

frame, id, x1, y1, x2, y2, score, class, visibility

此外，提取的数据中的x2和y2应该像这样：

x2_new = x2 + x1 = 630 + 602 = 1232
y2_new = y2 + y1 = 564 + 549 = 1113

预期的输出（假设提供的字典位于df1的第一行中）：

1054, 78, 602, 549, 1232, 1113, 1, 5, 0.0

我已经尝试过使用此代码在detail列的基础上创建新的df，但出现错误：

for i in finaldftoconvert['id']:
    for k in finaldftoconvert[['detail'][['id']==i]]:
        df = pd.DataFrame(k)
print df

错误：

main.py:267: FutureWarning: elementwise comparison failed; returning scalar instead, but in the future will perform elementwise comparison
  for k in finaldftoconvert[['detail'][['id']==i]]:
Traceback (most recent call last):
  File "main.py", line 268, in <module>
    df = pd.DataFrame(k)
  File "/usr/lib/python2.7/dist-packages/pandas/core/frame.py", line 305, in __init__
    raise PandasError('DataFrame constructor not properly called!')
pandas.core.common.PandasError: DataFrame constructor not properly called!

Answer 1

a = pd.DataFrame(index=[78],columns=["detail"])
a.loc[78,"detail"] = [{'y1': 549, 'score': 1, 'x2': 630, 'frame': 1054, 'y2': 564, 'x1': 602, 'visibility': 0.0, 'class': 5}]
a.loc[188,"detail"] = [{'y1': 649, 'score': 1, 'x2': 630, 'frame': 1054, 'y2': 564, 'x1': 602, 'visibility': 0.0, 'class': 5}]

对于每个字典，应使用pd.DataFrame.from_dict。实际上，我不是完全要打印出来吗？或将其转换为多个数据帧。

这是一些简单的解决方案。

# print it 
a.applymap(lambda x:print(pd.DataFrame.from_dict({0:x[0]})))
# convert it
for i in a.index:
    tmp = pd.DataFrame.from_dict({0:a.loc[i,"detail"][0]}).T
    tmp.x2 = tmp.x2+tmp.x1
    tmp.y2 = tmp.y2 + tmp.y1
    # this function you could storge in any dict/list. Or you could storge into a list. And using pd.concat to concate them together.

将数据框中的字典列表转换为CSV

1 个答案: