1 个回答
-
假设数据框的大小相同,则可以将分配
RESULT_df['RESULT'].values
给原始数据框。这样,您不必担心索引问题。# pre 0.24 feature_file_df['RESULT'] = RESULT_df['RESULT'].values # >= 0.24 feature_file_df['RESULT'] = RESULT_df['RESULT'].to_numpy()
最少的代码样本
df A B 0 -1.202564 2.786483 1 0.180380 0.259736 2 -0.295206 1.175316 3 1.683482 0.927719 4 -0.199904 1.077655 df2 C 11 -0.140670 12 1.496007 13 0.263425 14 -0.557958 15 -0.018375
让我们先尝试直接分配。
df['C'] = df2['C'] df A B C 0 -1.202564 2.786483 NaN 1 0.180380 0.259736 NaN 2 -0.295206 1.175316 NaN 3 1.683482 0.927719 NaN 4 -0.199904 1.077655 NaN
现在,分配由返回的数组
.values
(或.to_numpy()
对于> 0.24的熊猫版本)。.values
返回一个numpy
没有索引的数组。df2['C'].values array([-0.141, 1.496, 0.263, -0.558, -0.018]) df['C'] = df2['C'].values df A B C 0 -1.202564 2.786483 -0.140670 1 0.180380 0.259736 1.496007 2 -0.295206 1.175316 0.263425 3 1.683482 0.927719 -0.557958 4 -0.199904 1.077655 -0.018375