我有工作代码可以实现我的目标，但是如您所见，它不是很优雅。我尝试使用 for循环编写它，但是我的编码知识是相对基础的。有什么大方的人可以帮助我简化代码并希望对其进行注释，以便我自己学习如何做吗？我知道这是一个很大的问题，但感谢您的帮助，谢谢！
最终目标是要生成一张要导出的表，其中包含每种拟合的A0和kobs值（以及它们各自的标准误差）。
然后将kobs值传递到另一个方程式以获得三个附加参数。在提供的数据中，[100]数据集被排除在最终分析之外（排除是由外部因素证明的）。我是新来的人，因此，如果我格式化错误或违反某些标准，请告诉我，我们将对其进行修复。
数据（从.csv文件复制）：
Time,[0]1,[0]2,[0]3,[1],[2.5],[6],[16],[40],[100]
0,1.008,,0.963,1.038,0.979,0.973,0.973,0.906,0.979
0,0.992,1.000,1.037,0.962,1.021,1.027,1.027,1.094,1.021
5,0.813,0.968,0.961,0.704,0.667,0.470,,,
5,0.861,0.971,0.913,0.713,0.645,0.512,0.353,0.306,0.351
10,0.820,0.868,0.888,0.613,0.407,0.262,,0.158,0.236
10,0.851,0.857,0.890,0.563,0.444,0.250,0.197,0.169,0.275
15,0.778,0.773,0.798,,,,0.154,0.145,0.204
15,0.778,0.752,0.894,0.552,0.308,0.184,0.109,0.146,0.238
20,0.610,0.727,0.806,0.441,0.247,0.180,0.114,0.143,0.269
20,0.747,0.784,0.806,0.426,0.257,0.176,0.138,0.116,0.345
30,,,,0.321,0.164,0.091,0.127,0.100,0.279
30,0.563,0.642,0.633,0.268,0.146,0.082,0.096,0.096,0.275
40,0.556,0.524,0.529,0.286,0.106,0.082,0.037,0.079,0.249
40,0.581,0.485,0.487,0.266,0.119,0.045,,,0.041
## Assign Time (x) data
t <- D1$Time
## Assign Response (y) data
R1 <- D1$`[0]1`
R2 <- D1$`[0]2`
R3 <- D1$`[0]3`
R4 <- D1$`[1]`
R5 <- D1$`[2.5]`
R6 <- D1$`[6]`
R7 <- D1$`[16]`
R8 <- D1$`[40]`
R9 <- D1$`[100]`
## Fit data
F1 <- nls(R1 ~ A1 * exp(-k1 * t), start = list(A1 = 1, k1 = 0.02))
P1 <- summary(F1)$parameters[,1:2]
F2 <- nls(R2 ~ A1 * exp(-k1 * t), start = list(A1 = 1, k1 = 0.02))
P2 <- summary(F2)$parameters[,1:2]
F3 <- nls(R3 ~ A1 * exp(-k1 * t), start = list(A1 = 1, k1 = 0.02))
P3 <- summary(F3)$parameters[,1:2]
F4 <- nls(R4 ~ A1 * exp(-k1 * t), start = list(A1 = 1, k1 = 0.02))
P4 <- summary(F4)$parameters[,1:2]
F5 <- nls(R5 ~ A1 * exp(-k1 * t), start = list(A1 = 1, k1 = 0.02))
P5 <- summary(F5)$parameters[,1:2]
F6 <- nls(R6 ~ A1 * exp(-k1 * t), start = list(A1 = 1, k1 = 0.02))
P6 <- summary(F6)$parameters[,1:2]
F7 <- nls(R7 ~ A1 * exp(-k1 * t), start = list(A1 = 1, k1 = 0.02))
P7 <- summary(F7)$parameters[,1:2]
F8 <- nls(R8 ~ A1 * exp(-k1 * t), start = list(A1 = 1, k1 = 0.02))
P8 <- summary(F8)$parameters[,1:2]
F9 <- nls(R9 ~ A1 * exp(-k1 * t), start = list(A1 = 1, k1 = 0.02))
P9 <- summary(F9)$parameters[,1:2]
## Assemble Table
SS <- c(colnames(D1)[2],colnames(D1)[3],colnames(D1)[4],colnames(D1)[5],colnames(D1)[6],colnames(D1)[7],colnames(D1)[8],colnames(D1)[9],colnames(D1)[10])
A0 <- c(P1[1,1],P2[1,1],P3[1,1],P4[1,1],P5[1,1],P6[1,1],P7[1,1],P8[1,1],P9[1,1])
SEA0 <- c(P1[1,2],P2[1,2],P3[1,2],P4[1,2],P5[1,2],P6[1,2],P7[1,2],P8[1,2],P9[1,2])
kobs <- c(P1[2,1],P2[2,1],P3[2,1],P4[2,1],P5[2,1],P6[2,1],P7[2,1],P8[2,1],P9[2,1])
SEkobs <- c(P1[2,2],P2[2,2],P3[2,2],P4[2,2],P5[2,2],P6[2,2],P7[2,2],P8[2,2],P9[2,2])
ExTab <- cbind(SS, A0, SEA0, kobs, SEkobs)
write_clip(ExTab)
conI <- c(0,0,0,0.5,1.5,4,12,35)
kobsA <- c(P1[2,1],P2[2,1],P3[2,1],P4[2,1],P5[2,1],P6[2,1],P7[2,1],P8[2,1])
kFit <- nls(kobsA ~ k0 + ((kin*conI)/(KI+conI)), start = list(k0 = 0.1, kin = 0.2, KI = 3))
summary(kFit)
我的意愿是使用for循环来减少重复，但是我无法编写工作循环。我还想消除##Assign Response (y) data部分，以便可以将代码应用于列名中具有不同浓度的通用数据集，但可以使用nls函数内的D1[2]代替R1产生错误。
答案 0 :(得分：1)
您可以这样开始：
d={'dog':'ani','cat':'ani','green':'color','blue':'color'}
df1['col2']=df1.col1.map(d)
df1
col1 col2
0 cat ani
1 cat ani
2 dog ani
3 green color
4 blue color