मेरे पास निम्न डेटा है। एक्सेल शीट से तालिका पढ़ी जाती है।

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NA, NA, NA), AS = c(NA, NA, NA, NA, 262.678895309165, NA, 254.335184123748, 
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NA, NA, NA, NA), AW = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, 270.731079462496, NA, 272.475652027079, NA, 271.630116391662, 
NA, 273.868936656244, NA, 277.452861420827, NA, 283.75827668541, 
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NA, NA)), row.names = c(NA, -37L), class = c("tbl_df", "tbl", 
"data.frame"))

क्या इस तालिका को निम्न प्रारूप में बदलने का कोई सदिश तरीका है? परिणाम दो कॉलम ('सेल' और 'वैल्यू') के साथ एक नया data.frame/data.table होना चाहिए। प्रत्येक सेल, जो मूल तालिका में NA नहीं है, रूपांतरित तालिका में एक पंक्ति होगी। कॉलम 'वैल्यू' की सामग्री मान है, कॉलम 'सेल' की सामग्री में मूल सेल से जुड़े कॉलम और पंक्ति की जानकारी शामिल होनी चाहिए (जैसे AC9)

structure(list(cell = c("AC13", "AC15", "AC17", "AC19", "AC21", 
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283.75827668541, 290.848857599992)), row.names = c(NA, -301L), class = "data.frame")

वर्तमान में मैं वांछित परिणाम प्राप्त करने के लिए दो नेस्टेड लूप का उपयोग करता हूं:

cols <- ncol(dt)
rows <- nrow(dt)

transformedDT <- data.frame(cell=character(), value=numeric())

for (c in 1:cols) {
  currentCol <- colnames(dt)[c]
  for (r in 1:rows) {
    curCell <- dt[[r, c]]
    if (!is.na(curCell)){
      transformedDT <- rbind(transformedDT, data.frame(cell=paste0(currentCol,r), value=curCell))
    }
  }
}

चूंकि मेरे पास प्रोसेस करने के लिए बहुत सारी फाइलें हैं, इसलिए निष्पादन समय में सुधार करने की जरूरत है।

एक data.table समाधान एक dplyr समाधान के लिए बेहतर है क्योंकि मैं अपने वर्कहॉर्स के रूप में data.table पर ध्यान केंद्रित करता हूं।

0
jojop2 2 सितंबर 2020, 12:55

2 जवाब

सबसे बढ़िया उत्तर

हम एक नए कॉलम में एक पंक्ति अनुक्रमणिका बना सकते हैं और डेटा को लंबे प्रारूप में प्राप्त कर सकते हैं। cell कॉलम बनाने के लिए कॉलम का नाम और रो इंडेक्स पेस्ट करें।

library(data.table)

setDT(df)[, row := seq_len(.N)]
df <- melt(df, na.rm = TRUE, id.vars = 'row', variable.name = 'cell')
df[, cell := paste0(cell, row)][, row := NULL]
df

#     cell    value
#  1: AC13 280.9054
#  2: AC15 278.1079
#  3: AC17 273.4773
#  4: AC19 271.4118
#  5: AC21 273.6381
# ---              
#297: AW17 271.6301
#298: AW19 273.8689
#299: AW21 277.4529
#300: AW23 283.7583
#301: AW25 290.8489

समतुल्य dplyr उत्तर होगा :

library(dplyr)

df %>%
  mutate(row = row_number()) %>%
  tidyr::pivot_longer(cols = -row, names_to = 'cell', values_drop_na = TRUE) %>%
  mutate(cell = paste0(cell, row)) %>%
  select(-row)
2
Ronak Shah 2 सितंबर 2020, 13:06

आधार R के reshape का उपयोग करना। कॉलम नामों में पहले से उपयुक्त उपसर्ग जोड़ें।

names(dat) <- paste0("value.", names(dat))
r <- reshape(transform(dat, id=rownames(dat)), timevar="cell", varying=names(dat), dir="l")

head(na.omit(r))  ## result
#       id cell    value
# 13.AC 13   AC 280.9054
# 15.AC 15   AC 278.1079
# 17.AC 17   AC 273.4773
# 19.AC 19   AC 271.4118
# 21.AC 21   AC 273.6381
# 23.AC 23   AC 272.5016
0
jay.sf 2 सितंबर 2020, 13:24