如何拆分列 |分成多列[重复]答案

【问题标题】：how to split column by | into multiple columns [duplicate]如何拆分列 |分成多列[重复]
【发布时间】：2013-12-21 19:24:27
【问题描述】：

在 R 中：我有一个多行但只有一列的数据框。每行都有一长串字符，周期性地用 |标记。每次有 | 时我都想拆分字符标记，这样就有很多列。

1995-01-01|33.399999999999999|40.299999999999997|35.399999999999999|35.0|37.200000000000003|23.399999999999999|23.199999999999999|47.399999999999999|49.200000000000003|49.200000000000003|48.100000000000001|42.299999999999997|58.200000000000003|17.399999999999999|50.700000000000003|5.2999999999999998|20.600000000000001|38.5|43.299999999999997 etc.

每个字符串都以日期开头，然后是对应城市的数字。变量名也列成一个字符串，需要用“.”隔开。标记。

date.abilene_tx.akron_oh.albany_ny.albuquerque_nm.allentown_pa.amarillo_tx.anchorage_ak.asheville_nc.atlanta_ga etc.

非常感谢任何帮助！

【问题讨论】：

strsplit 可能，但是您是如何将其放入 R 中的？如果您从文件中读取它，您可能想查看read.table 的sep 参数。

标签： regex r dataframe strsplit

【解决方案1】：

你应该已经用这个命令从文件中加载了数据：

 dat <- read.table(filename, sep="|")

这将处理用“|”分隔的行但是你又说“字符串”用“。”分隔，所以如果这些以某种方式混合在 htat 文本文件中，您可能需要先使用readLines() 对输入进行一些预处理。

【讨论】：

【解决方案2】：

这是一个包含一列和 10 行的 data.frame，可能与您的相似：

dat <- "1995-01-01|33.399999999999999|40.299999999999997|35.399999999999999|35.0|37.200000000000003|23.399999999999999|23.199999999999999|47.399999999999999|49.200000000000003|49.200000000000003|48.100000000000001|42.299999999999997|58.200000000000003|17.399999999999999|50.700000000000003|5.2999999999999998|20.600000000000001|38.5|43.299999999999997 "

df <- data.frame(col1 = rep(dat, 10))

这是具有基于拆分 Col1 的新列的 data.frame：

foo <- data.frame(do.call('rbind', strsplit(as.character(df$col1),'|',fixed=TRUE)))
foo

           X1                 X2                 X3                 X4   X5                 X6
1  1995-01-01 33.399999999999999 40.299999999999997 35.399999999999999 35.0 37.200000000000003
2  1995-01-01 33.399999999999999 40.299999999999997 35.399999999999999 35.0 37.200000000000003
3  1995-01-01 33.399999999999999 40.299999999999997 35.399999999999999 35.0 37.200000000000003
4  1995-01-01 33.399999999999999 40.299999999999997 35.399999999999999 35.0 37.200000000000003
5  1995-01-01 33.399999999999999 40.299999999999997 35.399999999999999 35.0 37.200000000000003
6  1995-01-01 33.399999999999999 40.299999999999997 35.399999999999999 35.0 37.200000000000003
7  1995-01-01 33.399999999999999 40.299999999999997 35.399999999999999 35.0 37.200000000000003
8  1995-01-01 33.399999999999999 40.299999999999997 35.399999999999999 35.0 37.200000000000003
9  1995-01-01 33.399999999999999 40.299999999999997 35.399999999999999 35.0 37.200000000000003
10 1995-01-01 33.399999999999999 40.299999999999997 35.399999999999999 35.0 37.200000000000003
                   X7                 X8                 X9                X10                X11
1  23.399999999999999 23.199999999999999 47.399999999999999 49.200000000000003 49.200000000000003
2  23.399999999999999 23.199999999999999 47.399999999999999 49.200000000000003 49.200000000000003
3  23.399999999999999 23.199999999999999 47.399999999999999 49.200000000000003 49.200000000000003
4  23.399999999999999 23.199999999999999 47.399999999999999 49.200000000000003 49.200000000000003
5  23.399999999999999 23.199999999999999 47.399999999999999 49.200000000000003 49.200000000000003
6  23.399999999999999 23.199999999999999 47.399999999999999 49.200000000000003 49.200000000000003
7  23.399999999999999 23.199999999999999 47.399999999999999 49.200000000000003 49.200000000000003
8  23.399999999999999 23.199999999999999 47.399999999999999 49.200000000000003 49.200000000000003
9  23.399999999999999 23.199999999999999 47.399999999999999 49.200000000000003 49.200000000000003
10 23.399999999999999 23.199999999999999 47.399999999999999 49.200000000000003 49.200000000000003
                  X12                X13                X14                X15                X16
1  48.100000000000001 42.299999999999997 58.200000000000003 17.399999999999999 50.700000000000003
2  48.100000000000001 42.299999999999997 58.200000000000003 17.399999999999999 50.700000000000003
3  48.100000000000001 42.299999999999997 58.200000000000003 17.399999999999999 50.700000000000003
4  48.100000000000001 42.299999999999997 58.200000000000003 17.399999999999999 50.700000000000003
5  48.100000000000001 42.299999999999997 58.200000000000003 17.399999999999999 50.700000000000003
6  48.100000000000001 42.299999999999997 58.200000000000003 17.399999999999999 50.700000000000003
7  48.100000000000001 42.299999999999997 58.200000000000003 17.399999999999999 50.700000000000003
8  48.100000000000001 42.299999999999997 58.200000000000003 17.399999999999999 50.700000000000003
9  48.100000000000001 42.299999999999997 58.200000000000003 17.399999999999999 50.700000000000003
10 48.100000000000001 42.299999999999997 58.200000000000003 17.399999999999999 50.700000000000003
                  X17                X18  X19                 X20
1  5.2999999999999998 20.600000000000001 38.5 43.299999999999997 
2  5.2999999999999998 20.600000000000001 38.5 43.299999999999997 
3  5.2999999999999998 20.600000000000001 38.5 43.299999999999997 
4  5.2999999999999998 20.600000000000001 38.5 43.299999999999997 
5  5.2999999999999998 20.600000000000001 38.5 43.299999999999997 
6  5.2999999999999998 20.600000000000001 38.5 43.299999999999997 
7  5.2999999999999998 20.600000000000001 38.5 43.299999999999997 
8  5.2999999999999998 20.600000000000001 38.5 43.299999999999997 
9  5.2999999999999998 20.600000000000001 38.5 43.299999999999997 
10 5.2999999999999998 20.600000000000001 38.5 43.299999999999997

【讨论】：

感谢您的帮助！但是出现了一条错误消息：非字符参数。那是因为它是一个数据框吗？我该如何解决这个问题？
我已经更新了处理数据框的答案
这非常有用。我是 R 新手，并且正在学习。因为我的数据集有几千行具有不同的值，所以我需要创建一个 for 循环（或函数）来对所有行重复此步骤？
你试过了吗？这个函数应该适用于您的数据框，假设每行中的长字符串可以分成相同数量的片段（这些片段中的内容无关紧要，为了方便起见，我只是复制了第一行）。您为什么不接受这个答案，然后 ask another question 包含您的数据框样本（使用 dput(head(mydata)) 获取样本以粘贴到您的问题中）。您的问题与您的实际用例越接近，答案就越相关。