loop regression when column names have special characters

Question

Dummy data shown below. I tried amending my loop regression code to account for spaces in the column names, but it still gives an error. Additionally, I'm not sure how to adjust .x to account for spaces in the independent variable column names too. The purpose of the code is to regress the first column against each of the subsequent columns one at a time. i.e. Canada Price ~ Global Price, then Canada Price ~ Canada Bond Price and so on.

lm.test <- map_dfr(
  set_names(names(df)),
  ~ tidy(lm(as.formula(paste("`",colnames(df[1]),"`","~", .x)),
            data = df)),
  .id = "predictor") 

Canada Price	Global Price	Canada 2Y Bond Price	US - Canada Inflation	Oil Price
-0.030661468	-0.000207995	0.084	-0.075	0.002116
-0.034269304	-0.140212141	0.363	-1.062	-0.06122
0.02371464	0.019987	-0.781	0.273	0.209895
-0.007971484	0.090597341	-0.221	0.068	-0.10657
0.005342716	0.060627149	0.151	0.234	-0.22191
-0.002210376	0.046010712	-0.106	0.13	0.073975
0.018338299	0.019799534	0.036	0.005	0.058091
0.000662471	-0.025356047	0.523	0.045	-0.03765
0.002124411	0.045979342	-0.433	0.282	0.365933
0.034191555	0.043211347	-0.08	0.147	0.113365
-0.011470069	-0.033773176	0.396	-0.037	-0.09753
0.007039847	0.050327089	-0.052	0.151	0.145487
-0.028872959	-0.004501371	0.324	-0.214	0.064282
0.009244373	-0.00092413	0.166	-0.066	0.076961
0.017448201	-0.010638258	-0.179	0.053	0.108548
-0.002379819	0.050111674	0.352	-0.161	-0.11261
-0.002103406	0.03094893	0.121	0.106	0.130575
0.019154969	0.08307375	0.051	0.149	0.041074
-0.00124318	-0.054526954	0.345	0.032	0.079687
-0.001241636	0.003319001	-0.289	0.219	0.100941

---

Edit

Data in dput format.

df1 <-
  structure(list(
    `Canada Price` = c(-0.030661468, -0.034269304, 0.02371464, -0.007971484, 
                       0.005342716, -0.002210376, 0.018338299, 0.000662471, 0.002124411, 
                       0.034191555, -0.011470069, 0.007039847, -0.028872959, 0.009244373, 
                       0.017448201, -0.002379819, -0.002103406, 0.019154969, -0.00124318, -0.001241636), 
    `Global Price` = c(-0.000207995, -0.140212141, 0.019987, 0.090597341, 
                       0.060627149, 0.046010712, 0.019799534, -0.025356047, 0.045979342,
                       0.043211347, -0.033773176, 0.050327089, -0.004501371, -0.00092413, 
                       -0.010638258, 0.050111674, 0.03094893, 0.08307375, -0.054526954, 0.003319001), 
    `Canada 2Y Bond Price` = c(0.084, 0.363, -0.781, -0.221, 0.151, -0.106, 
                               0.036, 0.523, -0.433, -0.08, 0.396, -0.052, 0.324, 0.166, -0.179, 0.352,
                               0.121, 0.051, 0.345, -0.289), 
    `US - Canada Inflation` = c(-0.075, -1.062, 0.273, 0.068, 
                                0.234, 0.13, 0.005, 0.045, 0.282, 0.147, -0.037, 0.151, -0.214, 
                                -0.066, 0.053, -0.161, 0.106, 0.149, 0.032, 0.219), 
    `Oil Price` = c(0.002116, -0.06122, 0.209895, -0.10657, -0.22191, 0.073975, 
                    0.058091, -0.03765, 0.365933, 0.113365, -0.09753, 0.145487, 0.064282, 
                    0.076961, 0.108548, -0.11261, 0.130575, 0.041074, 0.079687, 0.100941)), 
    class = "data.frame", row.names = c(NA, -20L))

Answer 1

Here is a solution to run regressions on all other columns but the first.
The formula is created with reformulate.

suppressPackageStartupMessages({
  library(magrittr)
  library(purrr)
})

# create the formula with 'value' as regressor,
# it will be the column name after pivoting to long format
fmla <- reformulate("value", response = names(df1)[1L] |> as.name())
# run the regressions
lm_fit_list <- df1 %>%
  tidyr::pivot_longer(-1L) %>%
  split(.$name) %>%
  map(\(df) lm(fmla, data = df))
  
# this gives some statistics
lm_smry_list <- lm_fit_list %>% map(summary)
lm_smry_list %>% map(coef)
#> $`Canada 2Y Bond Price`
#>                 Estimate  Std. Error    t value   Pr(>|t|)
#> (Intercept)  0.001736073 0.003667346  0.4733866 0.64162781
#> value       -0.025789455 0.011690588 -2.2060015 0.04061964
#> 
#> $`Global Price`
#>                 Estimate  Std. Error    t value   Pr(>|t|)
#> (Intercept) -0.001515084 0.003712875 -0.4080622 0.68804100
#> value        0.164831106 0.070018144  2.3541199 0.03012634
#> 
#> $`Oil Price`
#>                 Estimate  Std. Error    t value  Pr(>|t|)
#> (Intercept) -0.001424249 0.004112478 -0.3463238 0.7331147
#> value        0.046411939 0.030404220  1.5264966 0.1442663
#> 
#> $`US - Canada Inflation`
#>                 Estimate  Std. Error    t value    Pr(>|t|)
#> (Intercept) 0.0001923356 0.003176245 0.06055441 0.952381362
#> value       0.0393945114 0.011325922 3.47826091 0.002683133

^{Created on 2024-02-23 with reprex v2.0.2}

---

Edit

After running summary on the output list, you can extract the values you want from this second list (of summaries).

suppressPackageStartupMessages({
  library(magrittr)
  library(purrr)
})

# create the formula with 'value' as regressor,
# it will be the column name after pivoting to long format
fmla <- reformulate("value", response = names(df1)[1L] |> as.name())
# run the regressions
lm_fit_list <- df1 %>%
  tidyr::pivot_longer(-1L) %>%
  split(.$name) %>%
  map(\(df) lm(fmla, data = df))
  
lm_smry_list <- lm_fit_list %>% map(summary)

stats <- lm_smry_list %>%
  map(coef) %>%
  map2(names(lm_smry_list), \(x, y) {
    data.frame(Variable = y) %>% cbind(x)
  }) %>% 
  list_rbind() 
row.names(stats) <- NULL
stats
#>                Variable      Estimate  Std. Error     t value    Pr(>|t|)
#> 1  Canada 2Y Bond Price  0.0017360725 0.003667346  0.47338661 0.641627809
#> 2  Canada 2Y Bond Price -0.0257894550 0.011690588 -2.20600149 0.040619640
#> 3          Global Price -0.0015150839 0.003712875 -0.40806217 0.688041004
#> 4          Global Price  0.1648311056 0.070018144  2.35411990 0.030126342
#> 5             Oil Price -0.0014242490 0.004112478 -0.34632381 0.733114717
#> 6             Oil Price  0.0464119390 0.030404220  1.52649664 0.144266337
#> 7 US - Canada Inflation  0.0001923356 0.003176245  0.06055441 0.952381362
#> 8 US - Canada Inflation  0.0393945114 0.011325922  3.47826091 0.002683133

^{Created on 2024-02-23 with reprex v2.0.2}

Answer 2

Map(\(x, y)coef(summary(lm(y~x))), df1[-1], df1[1])

$`Canada 2Y Bond Price`
                           Estimate  Std. Error    t value   Pr(>|t|)
(Intercept)             0.001736073 0.003667346  0.4733866 0.64162781
`Canada 2Y Bond Price` -0.025789455 0.011690588 -2.2060015 0.04061964

$`Global Price`
                   Estimate  Std. Error    t value   Pr(>|t|)
(Intercept)    -0.001515084 0.003712875 -0.4080622 0.68804100
`Global Price`  0.164831106 0.070018144  2.3541199 0.03012634

$`Oil Price`
                Estimate  Std. Error    t value  Pr(>|t|)
(Intercept) -0.001424249 0.004112478 -0.3463238 0.7331147
`Oil Price`  0.046411939 0.030404220  1.5264966 0.1442663

$`US - Canada Inflation`
                            Estimate  Std. Error    t value    Pr(>|t|)
(Intercept)             0.0001923356 0.003176245 0.06055441 0.952381362
`US - Canada Inflation` 0.0393945114 0.011325922 3.47826091 0.002683133

---

Note that if you are only interested in the coefficients ie prediction, and nothing to do with hypothesis testing, you could simply run one linear regression:

a <- lm(`Canada Price`~ind/values+0, cbind(df1[1], stack(df1, -1)))
as.matrix(coef(a))
                                         [,1]
indGlobal Price                 -0.0015150839
indCanada 2Y Bond Price          0.0017360725
indUS - Canada Inflation         0.0001923356
indOil Price                    -0.0014242490
indGlobal Price:values           0.1648311056
indCanada 2Y Bond Price:values  -0.0257894550
indUS - Canada Inflation:values  0.0393945114
indOil Price:values              0.0464119390

ie where you see :value that is the slope and the other is the intercept. eg

intercept and slope for Global Price are -0.0015150839 and 0.1648311056 respectively. Compare results to the lapply version