In the dataframe posted in question, there are NA values because there is not enough data to compute standard deviation. See below:
> df1<-data.frame("a"=0,"b"=1)
> df1%>%
+ summarize_all(funs( min , max ,mean, sd))
a_min b_min a_max b_max a_mean b_mean a_sd b_sd
1 0 1 0 1 0 1 NaN NaN
> df2<-data.frame("a"=c(0,1,2,3),"b"=c(1,3,5,7))
> df2%>%
+ summarize_all(funs( min , max ,mean, sd))
a_min b_min a_max b_max a_mean b_mean a_sd b_sd
1 0 1 3 7 1.5 4 1.290994 2.581989
In case you have NA values in the dataset, using na.rm=T will solve your purpose:
> df3<-data.frame("a"=c(0,1,NA,3),"b"=c(1,3,5,7))
# with na.rm=T
> df3%>%
+ summarize_all(funs( min , max ,mean, sd),na.rm=T)
a_min b_min a_max b_max a_mean b_mean a_sd b_sd
1 0 1 3 7 1.333333 4 1.527525 2.581989
# without na.rm=T
> df3%>%
+ summarize_all(funs( min , max ,mean, sd))
a_min b_min a_max b_max a_mean b_mean a_sd b_sd
1 NA 1 NA 7 NA 4 NaN 2.581989