python - Dataframe 从列表中添加多列，并创建每个列名

Question

有一个字典列表d，其中x是一个嵌入列表，例如

d = [{"name":"Python", "x":[0,1,2,3,4,5]},  # x has 300 elements
     {"name":"C++", "x":[0,1,0,3,4,4]},
     {"name":"Java","x":[0,4,5,6,1]}]

我想将 d 转换为 Dataframe，并为 x 中的每个元素自动添加列，添加的列名有一个前缀 "abc”，例如，

df.columns = ["name", "abc0", "abc1", ..., "abc300"]

我正在寻找一种有效的方法，因为 d 有很多字典。当我手动添加列时，Python 说

PerformanceWarning: DataFrame is highly fragmented.  This is usually the result of calling `frame.insert` many times, which has poor performance.  Consider joining all columns at once using pd.concat(axis=1) instead.  To get a de-fragmented frame, use `newframe = frame.copy()`

Answer 1

你在找这样的东西吗:

d = [{"name":"Python", "x":[0,1,2,3,4,5]},  # x has 300 elements
     {"name":"C++", "x":[0,1,0,3,4,4]},
     {"name":"Java","x":[0,4,5,6,1]}]

df = pd.DataFrame(
    {
        "name": record["name"],
        **{f"abc{i}": n for i, n in enumerate(record["x"])}
    }
    for record in d
)

您的样本结果:

     name  abc0  abc1  abc2  abc3  abc4  abc5
0  Python     0     1     2     3     4   5.0
1     C++     0     1     0     3     4   4.0
2    Java     0     4     5     6     1   NaN