如何从文本中识别位置

本文介绍了如何从文本中识别位置的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 df = read.csv（secondary.csv，header = TRUE ） 解决方案 S < - S / O SK hungu九十零分之一百〇一MODEL HOUSE TALAB Gagni酒店SHUKUL LUCKNOW北方邦LUCKNOW北方邦226001 我建议制作所有可能的Nx个字符串，其中N是字符串的长度，x是可变长度的 allchr< - unset（strsplit（S，）） listsubstr ＃[1] S / O SK hungu九十分之一百○一MODEL HOUSE TALAB Gagni酒店SHUKUL LUCKNOW北方邦LUCKNOW北方邦226001 ＃[2]/ O SK hungu九十分之一百○一MODEL HOUSE TALAB Gagni酒店SHUKUL LUCKNOW北方邦LUCKNOW北方邦226001\" ＃[3] / O SK hungu九十零分之一百零一MODEL HOUSE TALAB Gagni酒店SHUKUL LUCKNOW北方邦LUCKNOW北方邦226001 ＃[4] O- SK hungu九十零分之一百零一MODEL HOUSE TALAB Gagni酒店SHUKUL LUCKNOW北方邦LUCKNOW北方邦226001 您可以遍历这个列表来检查有效的地理编码。我不得不提供伪代码，因为我不知道如何检查一个字符串是否是有效的地理编码。 pre $ sapply（listsubstr，函数（I）is.geocode（I））＃包含伪代码 你也可以用递归（x是gecode）{＃包含伪代码$（$） $ b myfun< b $ b return（x）} else { myfun（substr（x，2，nchar（S）））} } Here is sample to my function that getscodes df= read.csv("secondary.csv",header = TRUE) 解决方案 S <- "s / O sk hungu 101 / 90 MODEL HOUSE TALAB GAGNI SHUKUL LUCKNOW UTTAR PRADESH LUCKNOW UTTAR PRADESH 226001"I recommend making all possible N-x strings where N is length of your string and x is variable lengthallchr <- unlist(strsplit(S, ""))listsubstr <- sapply(1:length(allchr), function(I) paste0(allchr[I:length(allchr)], collapse="")) # [1] "s / O sk hungu 101 / 90 MODEL HOUSE TALAB GAGNI SHUKUL LUCKNOW UTTAR PRADESH LUCKNOW UTTAR PRADESH 226001" # [2] " / O sk hungu 101 / 90 MODEL HOUSE TALAB GAGNI SHUKUL LUCKNOW UTTAR PRADESH LUCKNOW UTTAR PRADESH 226001" # [3] "/ O sk hungu 101 / 90 MODEL HOUSE TALAB GAGNI SHUKUL LUCKNOW UTTAR PRADESH LUCKNOW UTTAR PRADESH 226001" # [4] " O sk hungu 101 / 90 MODEL HOUSE TALAB GAGNI SHUKUL LUCKNOW UTTAR PRADESH LUCKNOW UTTAR PRADESH 226001"You can iterate through this list to check for valid geocodes. I have to provide pseudocode since I'm not sure how to check if a string is a valid geocode.sapply(listsubstr, function(I) is.geocode(I)) # contains pseudocodeYou could also do this with recursion though.myfun <- function(x) { if (x is gecode) { # contains pseudocode return(x) } else { myfun(substr(x, 2, nchar(S))) } } 这篇关于如何从文本中识别位置的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！