我需要通過索引將 pd.DataFrame 中的一堆行盡可能高效地分配給已知大小的矩陣。如果索引在資料框中不存在，則很多，并且這些行必須用 NaN 填充。此操作將在迭代資料幀中的每一行的回圈內發生，并且必須盡可能快。
簡而言之，我想加快以下順序：
# DF to iterate over
df = pd.DataFrame({'a': [1,2,3,4], 'b': [2,4,6,8], 'c': [3,6,9,12]})
# (Relative) DF indices to return if possible
indices = [-2, 0, 2, 4]
len_df = len(df)
len_indices = len(indices)
len_cols = len(df.columns)
values = np.zeros((len_df, len_indices, len_cols))
for i in range(len_df):
for j, n in enumerate(indices):
idx = i n
try:
assert idx > 0 # avoid wrapping around
values[i, j] = df.iloc[idx]
except:
values[i, j] = np.nan
values
退貨
[[[nan, nan, nan],
[ 1, 2, 3],
[ 3, 6, 9],
[nan, nan, nan]],
[[nan, nan, nan],
[ 2, 4, 6],
[ 4, 8, 12],
[nan, nan, nan]],
[[ 1, 2, 3],
[ 3, 6, 9],
[nan, nan, nan],
[nan, nan, nan]],
[[ 2, 4, 6],
[ 4, 8, 12],
[nan, nan, nan],
[nan, nan, nan]]]
根據需要，但速度很慢。有什么建議？
如預期的
uj5u.com熱心網友回復：
這是構建查找表的解決方案：
# repeat indices into new dimension
rep_ids = np.tile(indices, (len(df), 1))
# add row index to the tiled ids
range_ids = rep_ids np.arange(len(df))[:, np.newaxis]
# check which ids make a valid lookup in the df
# TODO: do you really mean > 0 here in the original code, not >= 0?
valid_test = (range_ids > 0) & (range_ids < len(df))
# lookup table, using 0 for invalid ids (will be overwritten later)
lookup_ids = np.where(valid_test, range_ids, 0)
# lookup all values
values = df.values[lookup_ids].astype(np.float64)
# set the invalids to nan
values[~valid_test] = np.nan
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