subset

Return values with matching conditions in r

由 半城伤御伤魂 提交于 2021-02-10 06:12:01
问题 I would like to return values with matching conditions in another column based on a cut score criterion. If the cut scores are not available in the variable, I would like to grab closest larger value. Here is a snapshot of dataset: ids <- c(1,2,3,4,5,6,7,8,9,10) scores.a <- c(512,531,541,555,562,565,570,572,573,588) scores.b <- c(12,13,14,15,16,17,18,19,20,21) data <- data.frame(ids, scores.a, scores.b) > data ids scores.a scores.b 1 1 512 12 2 2 531 13 3 3 541 14 4 4 555 15 5 5 562 16 6 6

Return values with matching conditions in r

由 人盡茶涼 提交于 2021-02-10 06:08:31
问题 I would like to return values with matching conditions in another column based on a cut score criterion. If the cut scores are not available in the variable, I would like to grab closest larger value. Here is a snapshot of dataset: ids <- c(1,2,3,4,5,6,7,8,9,10) scores.a <- c(512,531,541,555,562,565,570,572,573,588) scores.b <- c(12,13,14,15,16,17,18,19,20,21) data <- data.frame(ids, scores.a, scores.b) > data ids scores.a scores.b 1 1 512 12 2 2 531 13 3 3 541 14 4 4 555 15 5 5 562 16 6 6

Return values with matching conditions in r

由 被刻印的时光 ゝ 提交于 2021-02-10 06:08:29
问题 I would like to return values with matching conditions in another column based on a cut score criterion. If the cut scores are not available in the variable, I would like to grab closest larger value. Here is a snapshot of dataset: ids <- c(1,2,3,4,5,6,7,8,9,10) scores.a <- c(512,531,541,555,562,565,570,572,573,588) scores.b <- c(12,13,14,15,16,17,18,19,20,21) data <- data.frame(ids, scores.a, scores.b) > data ids scores.a scores.b 1 1 512 12 2 2 531 13 3 3 541 14 4 4 555 15 5 5 562 16 6 6

Return values with matching conditions in r

由 有些话、适合烂在心里 提交于 2021-02-10 06:08:25
问题 I would like to return values with matching conditions in another column based on a cut score criterion. If the cut scores are not available in the variable, I would like to grab closest larger value. Here is a snapshot of dataset: ids <- c(1,2,3,4,5,6,7,8,9,10) scores.a <- c(512,531,541,555,562,565,570,572,573,588) scores.b <- c(12,13,14,15,16,17,18,19,20,21) data <- data.frame(ids, scores.a, scores.b) > data ids scores.a scores.b 1 1 512 12 2 2 531 13 3 3 541 14 4 4 555 15 5 5 562 16 6 6

[LeetCode] 851. Loud and Rich 聒噪与富有

由 老子叫甜甜 提交于 2021-02-09 00:11:27
<br> In a group of N people (labelled `0, 1, 2, ..., N-1`), each person has different amounts of money, and different levels of quietness. For convenience, we'll call the person with label x , simply "person x ". We'll say that richer[i] = [x, y] if person x definitely has more money than person y . Note that richer may only be a subset of valid observations. Also, we'll say quiet[x] = q if person x has quietness q . Now, return answer , where answer[x] = y if y is the least quiet person (that is, the person y with the smallest value of quiet[y] ), among all people who definitely have equal to

Unable to subset (filter) a data frame due to NA's

由 China☆狼群 提交于 2021-02-08 19:12:12
问题 Why in the code below dplyr's filter doesn't return the same data.frame as base R subsetting? In fact none of them works as expected. I'd like to remove observations/rows which, simultaneously, b==1 AND c==1 . That is, I'd like to remove only the third row. require(dplyr) df <- data.frame(a=c(0,0,0,0,1,1,1), b=c(0,0,1,1,0,0,1), c=c(1,NA,1,NA,1,NA,NA)) filter(df, !(b==1 & c==1)) df[!(df$b==1 & df$c==1),] 回答1: Or use complete.cases to convert NA to FALSE in the result logic vector so that you

Unable to subset (filter) a data frame due to NA's

由 僤鯓⒐⒋嵵緔 提交于 2021-02-08 19:10:57
问题 Why in the code below dplyr's filter doesn't return the same data.frame as base R subsetting? In fact none of them works as expected. I'd like to remove observations/rows which, simultaneously, b==1 AND c==1 . That is, I'd like to remove only the third row. require(dplyr) df <- data.frame(a=c(0,0,0,0,1,1,1), b=c(0,0,1,1,0,0,1), c=c(1,NA,1,NA,1,NA,NA)) filter(df, !(b==1 & c==1)) df[!(df$b==1 & df$c==1),] 回答1: Or use complete.cases to convert NA to FALSE in the result logic vector so that you

Unable to subset (filter) a data frame due to NA's

由 淺唱寂寞╮ 提交于 2021-02-08 19:06:34
问题 Why in the code below dplyr's filter doesn't return the same data.frame as base R subsetting? In fact none of them works as expected. I'd like to remove observations/rows which, simultaneously, b==1 AND c==1 . That is, I'd like to remove only the third row. require(dplyr) df <- data.frame(a=c(0,0,0,0,1,1,1), b=c(0,0,1,1,0,0,1), c=c(1,NA,1,NA,1,NA,NA)) filter(df, !(b==1 & c==1)) df[!(df$b==1 & df$c==1),] 回答1: Or use complete.cases to convert NA to FALSE in the result logic vector so that you

Choosing only a monthly subset using quantmod

由 ◇◆丶佛笑我妖孽 提交于 2021-02-08 09:51:29
问题 I'm using the latest 5 years of apple and google using the quant mode command loadSymbols(c("AAPL", "GOOG")) AAPL['2016::'] This is the dataset. AAPL.Open AAPL.High AAPL.Low AAPL.Close AAPL.Volume AAPL.Adjusted 2016-01-04 25.6525 26.3425 25.5000 26.3375 270597600 24.36454 2016-01-05 26.4375 26.4625 25.6025 25.6775 223164000 23.75398 2016-01-06 25.1400 25.5925 24.9675 25.1750 273829600 23.28912 2016-01-07 24.6700 25.0325 24.1075 24.1125 324377600 22.30621 2016-01-08 24.6375 24.7775 24.1900 24

R: “Filter” columns in a data frame by a list [closed]

由 别来无恙 提交于 2021-02-08 07:46:31
问题 Closed. This question needs debugging details. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 4 years ago . Improve this question I want to produce a new data frame from my existing one, where the columns in this new df are selected based on whether that variable is listed in a separate vector (i.e., as rows). The new df would therefore only contain those columns that were listed in the vector. I want to do
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