包含以下內容的數據框 0 1 2 3 4 5 0 reviewers#0 -0.016271 0.011541 0.011903 -0.001355 0.0087021 reviewers#1 -0.037961 0.033415 0.020643 0.004748 0.0145232 reviewers#2 -0.019863 0.019906 0.017248 0.006216 0.0088263 reviewers#3 -0.021029 0.016401 0.010772 0.001874 0.0057724 reviewers#4 -0.013409 0.011703 0.011249 0.000111 0.0093195 reviewers#5 -0.008549 0.007816 0.007859 0.000984 0.0054916 reviewers#6 -0.01634 0.017007 0.014637 0.005241 0.0081357 reviewers#7 -0.017075 0.016119 0.013666 0.00314 0.0087868 reviewers#8 -0.030823 0.020217 0.012402 -0.003165 0.0096439 reviewers#9 -0.038311 0.026252 0.017619 0.003568 0.00397210 confiscated#0 -0.007147 0.007387 0.010867 0.000735 0.01124411 confiscated#1 -0.016917 0.014412 0.016182 0.001859 0.01559612 confiscated#2 -0.004854 0.004091 0.005075 -0.000566 0.0045813 confiscated#3 -0.02642 0.021311 0.018871 -0.001843 0.01703314 confiscated#4 -0.016161 0.013325 0.013113 -0.001036 0.01138515 confiscated#5 -0.0131 0.0117 0.013829 -0.000861 0.0122516 confiscated#6 -0.006454 0.005335 0.006634 -0.001038 0.00632217 confiscated#7 -0.006855 0.005225 0.007626 -0.003071 0.00904818 confiscated#8 -0.019227 0.015683 0.016805 -0.004709 0.01945319 confiscated#9 -0.010685 0.011237 0.011653 0.003006 0.007464
1 回答
胡子哥哥
TA貢獻1825條經驗 獲得超6個贊
聽起來這就是你所追求的：
import pandas as pd
from collections import defaultdict
df = pd.DataFrame([
{0: 'reviewers#0', 1: 1.0, 2: 2.0, 3: 3.0},
{0: 'reviewers#1', 1: 4.0, 2: 5.0, 3: 6.0},
{0: 'confiscated#0', 1: 7.0, 2: 8.0, 3: 9.0},
{0: 'confiscated#1', 1: 10.0, 2: 11.0, 3: 12.0},
])
print(df)
result = defaultdict(list)
for _, values in df.iterrows():
values = list(values)
result[int(values[0].split('#')[1])] += [values[1:]]
print(result)
結果：
0 1 2 3
0 reviewers#0 1.0 2.0 3.0
1 reviewers#1 4.0 5.0 6.0
2 confiscated#0 7.0 8.0 9.0
3 confiscated#1 10.0 11.0 12.0
defaultdict(<class 'list'>, {0: [[1.0, 2.0, 3.0], [7.0, 8.0, 9.0]], 1: [[4.0, 5.0, 6.0], [10.0, 11.0, 12.0]]})
即數據框具有您描述的結構，result字典是您需要的結果嗎？
如果您需要defaultdict成為一個簡單的字典，則需要進行一些小的更改，或者您可以將其轉換 - 但這不應該是一個問題。
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