tmp=rowSums(df[,-1])
df[is.na(tmp) | tmp!=0,]

              Shoes q1_sales q2_sales q3_sales q4_sales
1 Winter Collection       NA       NA       NA       NA
2             Boots       50       25        0      100
3          Trainers       40       65      200      100
4           Loafers        0       45       20       10
5 Summer Collection       NA       NA       NA       NA
6          Trainers       50      655      400        0
7        Flip Flops        0       25       40        0
8           Sandals        0      205       45       25
9             Other       NA       NA       NA       NA

对于dplyr，您可以在filter()中使用if_all和if_any：

library(dplyr)

df %>%
  filter(!if_all(-Shoes, ~ .x == 0) | if_any(-Shoes, is.na))

#               Shoes q1_sales q2_sales q3_sales q4_sales
# 1 Winter_Collection       NA       NA       NA       NA
# 2             Boots       50       25        0      100
# 3          Trainers       40       65      200      100
# 4           Loafers        0       45       20       10
# 5 Summer_Collection       NA       NA       NA       NA
# 6          Trainers       50      655      400        0
# 7        Flip_Flops        0       25       40        0
# 8           Sandals        0      205       45       25
# 9             Other       NA       NA       NA       NA

数据

df <- structure(list(Shoes = c("Winter_Collection", "Boots", "Trainers", 
"Loafers", "Summer_Collection", "Trainers", "Flip_Flops", "Sandals", 
"Other", "Clogs"), q1_sales = c(NA, 50L, 40L, 0L, NA, 50L, 0L, 
0L, NA, 0L), q2_sales = c(NA, 25L, 65L, 45L, NA, 655L, 25L, 205L, NA, 0L),
q3_sales = c(NA, 0L, 200L, 20L, NA, 400L, 40L, 45L, NA, 0L),
q4_sales = c(NA, 100L, 100L, 10L, NA, 0L, 0L, 25L, NA, 0L)),
class = "data.frame", row.names = c(NA, -10L))

• EDIT*：（跟随@DarrenTsai的宝贵评论）

tidyverse解决方案：

df %>%
  filter(!if_all(matches("^q"), ~ . == 0) | if_any(matches("^q"), is.na))