我有一个数据框，它是-> https://drive.google.com/file/d/1qcQRwmFIkTJHPaknXjV1vNlDScw1Fxf6/view?usp=sharing
Kyphosis Age Number Start prob_Age prob_Number prob_Start
50 absent 68 5 10 0.993964 0.208729 0.916693
51 absent 9 2 17 0.997321 0.904427 0.047178
52 present 139 10 6 0.004772 0.001366 0.964974
53 absent 2 2 17 0.997710 0.904427 0.047178
54 absent 140 4 15 0.004711 0.779213 0.072759
55 absent 72 5 15 0.993830 0.208729 0.072759
56 absent 2 3 13 0.997710 0.829827 0.090356
57 present 120 5 8 0.005786 0.208729 0.939803
58 absent 51 7 9 0.994754 0.072175 0.927241
59 absent 102 3 13 0.006362 0.829827 0.090356
60 present 130 4 1 0.005290 0.779213 0.996493
61 present 114 7 8 0.006029 0.072175 0.939803
62 absent 81 4 1 0.993617 0.779213 0.996493
63 absent 118 3 16 0.005872 0.829827 0.060197
64 absent 118 4 16 0.005872 0.779213 0.060197
65 absent 17 4 10 0.996844 0.779213 0.916693
66 absent 195 2 17 0.001558 0.904427 0.047178
67 absent 159 4 13 0.003517 0.779213 0.090356
68 absent 18 4 11 0.996783 0.779213 0.909644
69 absent 15 5 16 0.996966 0.208729 0.060197
70 absent 158 5 14 0.003580 0.208729 0.083307
71 absent 127 4 12 0.005449 0.779213 0.092836
72 absent 87 4 16 0.993547 0.779213 0.060197
73 absent 206 4 10 0.001135 0.779213 0.916693
74 absent 11 3 15 0.997205 0.829827 0.072759
75 absent 178 4 15 0.002387 0.779213 0.072759
76 present 157 3 13 0.003643 0.829827 0.090356
77 absent 26 7 13 0.996282 0.072175 0.090356
78 absent 120 2 13 0.005786 0.904427 0.090356
79 present 42 7 6 0.995277 0.072175 0.964974
80 absent 36 4 13 0.995648 0.779213 0.090356
我有这些列表：
A=0,S=0,N=0
X3=[A,S]
X7=[N,A,A,A,S,S]
X5=[S,N,A,A,S,A,S]
X4=[N,S,N,A,A,S,A,S]
X9=[N,S,N,A,A,S,A,S]
X10=[A,A,A,S,S]
list=[ X7, X7, X5, X7, X7, X7, X7, X5, X7, X7, X5, X5, X3, X7, X7, X7, X10, X10, X7, X7, X10, X7, X7, X10, X7, X10, X9, X7, X7, X4,X7]
现在，我的目标是通过df，将column的值放入 每条记录的“ prob_Age”，“ prob_Number”，“ prob_Start”分别放入“列表”
我尝试了以下代码：
A=0,S=0,N=0
X3=[N,A,S]
X7=[A,S]
X5=[A,S]
X4=[N,A,S]
X9=[A,A,S]
X10=[A,A,S]
list=[ X7, X7, X5, X7, X7, X7, X7, X5, X7, X7, X5, X5, X3, X7, X7, X7, X10, X10, X7, X7, X10, X7, X7, X10, X7, X10, X9, X7, X7, X4,X7]
list1=[]
for i in df.iterrows():
A=df['prob_Age']
S=df['prob_Number']
N=df['prob_Start']
print(list)
预期产量
list=[ [0.993964,0.916693], [0.997321,0.047178], [0.004772,0.964974], [0.997710,0.047178], [0.004711,0.072759],
[0.993830,0.072759], [0.997710,0.090356], [0.005786,0.939803], [0.994754,0.927241], [0.006362,0.090356],
[0.005290,0.996493], [0.006029,0.939803], [0.993617,0.779213,0.996493], [0.005872,0.060197], [0.005872,0.060197],
[0.996844,0.916693], [0.001558,0.001558,0.047178], [0.003517,0.090356], [ 0.996783,0.909644], [0.996966,0.060197],
[0.003580,0.003580,0.083307], [0.005449,0.092836], [0.993547,0.060197], [0.001135,0.001135,0.916693], [0.997205,0.072759],
[ 0.002387,0.002387,0.072759], [0.003643,0.003643 ,0.090356], [0.996282 ,0.090356], [0.005786,0.090356], [0.995277,0.072175,0.964974],[0.995648,0.090356]]
我得到了答案，谢谢大家：
list=[]
c=0
for _, x in df.iterrows():
A, S, N = x[['prob_Age', 'prob_Start', 'prob_Number']].values
X3=[N,A,S]
X7=[A,S]
X5=[A,S]
X4=[N,A,S]
X9=[A,A,S]
X10=[A,A,S]
l=[ X7, X7, X5, X7, X7, X7, X7, X5, X7, X7, X5, X5, X3, X7, X7, X7, X10, X10, X7, X7, X10, X7, X7, X10, X7, X10, X9, X7, X7, X4, X7]
list.append(l[c])
c=c+1
print(list)
答案 0 :(得分：1)
首先，list是python中的内置函数，因此您不应真正将其用作变量名。其次，尽管您在每次迭代中都更改了变量A，S，N（但实际上并没有改变，因为在每次迭代中您都为其分配了相同的值），但是您并没有更改任何列表的值。因此，要获得每次迭代所需的输出，您应该执行以下操作：
for _, x in df.iterrows():
A, S, N = x[['prob_Age', 'prob_Number', 'prob_Start']].values
X3=[N,A,S]
X7=[A,S]
X5=[A,S]
X4=[N,A,S]
X9=[A,A,S]
X10=[A,A,S]
l=[ X7, X7, X5, X7, X7, X7, X7, X5, X7, X7, X5, X5, X3, X7, X7, X7, X10, X10, X7, X7, X10, X7, X7, X10, X7, X10, X9, X7, X7, X4,X7]
print(l)
现在，根据您的最终目标是什么，我很确定有更好的解决方案。
编辑 这可能会更好一点：
inds = [
'X7', 'X7', 'X5', 'X7', 'X7', 'X7', 'X7', 'X5', 'X7', 'X7',
'X5', 'X5', 'X3', 'X7', 'X7', 'X7', 'X10', 'X10', 'X7', 'X7',
'X10', 'X7', 'X7', 'X10', 'X7', 'X10', 'X9', 'X7', 'X7', 'X4', 'X7'
]
def fill_in(idx, row):
A, S, N = row[['prob_Age', 'prob_Number', 'prob_Start']].values
d = {
'X3': [N,A,S],
'X7': [A,S],
'X5': [A,S],
'X4': [N,A,S],
'X9': [A,A,S],
'X10': [A,A,S]
}
return d[inds[idx]]
l = [fill_in(i, x) for i, x in df.iterrows()]
答案 1 :(得分：0)
这可能不是最好的方法。
#A=0,S=0,N=0 This is not required
#X3=[N,A,S]
#X7=[A,S]
#X5=[A,S]
#X4=[N,A,S]
#X9=[A,A,S]
#X10=[A,A,S]
l =[ 'X7', 'X7', 'X5', 'X7', 'X7', 'X7', 'X7', 'X5'........]
# as pointed out above list should not be used as a variable
# change l values into string, there are easy ways to do this.
idx = 0
for i,row in df.iterrows(): # have to add the row because there are two values index and row to unpack
A=row['prob_Age']
S=row['prob_Number']
N=row['prob_Start']
X3=[N,A,S]
X7=[A,S]
X5=[A,S]
X4=[N,A,S]
if str(list[idx]) == 'X7': # the str part is not required if the l is changed.
list[idx] = X7
elif str(list[idx]) == 'X3':
list[idx] = X3
elif str(list[idx]) == 'X5':
list[idx] = X5
elif str(list[idx]) == 'X4':
list[idx] = X4
### Put more conditions
idx +=1
答案 2 :(得分：0)
问题尚不完全清楚，但是如果我理解正确，并且希望每行包含每个列表的列表（val 1，val 2，val 3），则此方法有效。我模拟了类似的df（示例中为“ {data}”：https://imgur.com/a/MhTuUbh）进行测试。
list_to_fill = []
length = len(data)
row = 0
col = 4
cell = data.iat[row, col]
for r in range (length):
temp_row_list = []
for i in range(3):
cell = data.iat[row, col]
temp_row_list.append(cell)
col = col + 1
list_to_fill.append(temp_row_list)
col = 4
row = row + 1
print ('final list =', list_to_fill)
给予您
最终列表= [[4，5，1]，[7，3，2]，[9，0，8]]