Question

当前，我有两个数据帧。第一个df1包含两列，代表网络联系。另一个df2仅包含一列，列出了我拥有属性数据的所有案例。

我想从df1中的df2中搜索那些案例，如果它们不在df1中的一列或两列中，那么我想删除该行。因此，我将得到一个df1，其中包含我拥有属性数据的案例之间的网络联系。

df1具有大约240万个关系（边缘列表），而df2具有3.4万个单独案例。

这是我在谷歌搜索一段时间后尝试执行的操作：

首先，我在df1上复制两列以进行备份。

library*dlypr)
df3<- df1%>%
     mutate(friendid = friendid %in% df2$V1,
            friendid = friendid*1.0) #converts boolean to numeric
df3<- df3%>%
     mutate(tieid= tieid%in% df2$V1,
            tieid= tieid*1.0)
#So what I think is going on here is that if the number is not found it turn the value into 0 and 1 if present. I do this for the two original columns from df1.

#Then I attempt to delete the rows by searching for 0 values on each column (2 and 3, which contain the edgelist).

df3<-df3[apply(df3[2],1,function(z) !any(z==0)),] 

df3<-df3[apply(df3[3],1,function(z) !any(z==0)),]

该过程确实删除了一堆行，但最终导致大约2k例。这似乎是不对的。我尝试了一种类似的方法来执行Excel，但是它对您一次可以加载的行数有限制。将数据集分为三个不同的文件并使用Kutools之后，我最终得到了约74k个案例。但是由于我做了很多手工工作，所以我可以肯定Excel工作中有错误。 R允许我一次加载所有数据，这将有助于获得更具体的结果。

任何帮助将不胜感激。谢谢

已编辑以提供更多信息：

> head(df1)
    ID     steamid    friendid daysknown    years         el1         el2
1    NA 7.65612e+16 7.65612e+16      2156 5.902806 7.65612e+16 7.65612e+16
2    NA 7.65612e+16 7.65612e+16      3480 9.527721 7.65612e+16 7.65612e+16
3    NA 7.65612e+16 7.65612e+16      1588 4.347707 7.65612e+16 7.65612e+16
4    NA 7.65612e+16 7.65612e+16       501 1.371663 7.65612e+16 7.65612e+16
5    NA 7.65612e+16 7.65612e+16       858 2.349076 7.65612e+16 7.65612e+16
6    NA 7.65612e+16 7.65612e+16       686 1.878166 7.65612e+16 7.65612e+16
> head(df2)
                 V1
1 76561197960265800
2 76561197960266000
3 76561197960266100
4 76561197960267800
5 76561197960268100
6 76561197960268400

df1中的Steamid和Friendid两列都必须是df2 $ V1中可用的ID。如果一对中仅存在一个ID，则该行也必须删除，如果也不存在。结束df将具有只能在df2中找到的ID对。

Answer 1

您可以执行以下操作：

df2$flag <- 1 #create a lookup column

df_temp <- merge(df1, df2, by.x = "friendid", by.y = "V1", all.x = TRUE)
names(df_temp) <- c("friendid", "tieid", "flag_1")
df_new <- merge(df_temp, df2, by.x = "tieid", by.y = "V1", all.x = TRUE)
names(df_temp) <- c("friendid", "tieid", "flag_1", "flag_2")

df_final <- subset(df_new, df_new$flag_1 == 1 | df_new$flag_2 == 1)

首先，您正在尝试检查df1和df2中匹配的朋友。然后，您要检查新数据帧和df2之间匹配的tieid。然后，您将子集新创建的数据框，以仅保留其中一个为1的行

Answer 2

您好Juan Juan Arroyo Flores，欢迎来到stackoverflow。

我不确定我是否正确，但是我认为您可以使用％in％运算符来解决此问题。

df $ variable1％in％df2 $ variable将检查df $ variable1中的每个元素（如果存在于df2 $ variable中）。

    df1 = data.frame("name1" = c("a", "b", "c", "d"), "name2" = c("f", "g", "h", "i"), stringsAsFactors = F)
    df2 = data.frame("names" = c("a", "g", "i"), stringsAsFactors = F)

    df1
    df2


    # name1 name2
    # 1     a     f
    # 2     b     g
    # 3     c     h
    # 4     d     i
    # > df2
    # names
    # 1     a
    # 2     g
    # 3     i

    # so we want to have row 1 selecet (cause of a), row 2 (caus of g) and row 4 (caus of i)
    # row 3 gets deleated

    # lets use %in% 

    df1$name1 %in% df2$names

    # > df1$name1 %in% df2$names
    # [1]  TRUE FALSE FALSE FALSE

    df1$name2 %in% df2$names

    # > df1$name2 %in% df2$names
    # [1] FALSE  TRUE FALSE  TRUE

    # to combine both a or is needed

    df1$name1 %in% df2$names | df1$name2 %in% df2$names

    # > df1$name1 %in% df2$names | df1$name2 %in% df2$names
    # [1]  TRUE  TRUE FALSE  TRUE

    # with which you can select the index 
    select_index = which(df1$name1 %in% df2$names | df1$name2 %in% df2$names)
    select_index

    # > select_index
    # [1] 1 2 4

    # now this can be used to select the desired rows
    df1[select_index,]

    # > df1[select_index,]
    # name1 name2
    # 1     a     f
    # 2     b     g
    # 4     d     i

    # you could as well just use 
    df1[df1$name1 %in% df2$names | df1$name2 %in% df2$names,]

    # > df1[df1$name1 %in% df2$names | df1$name2 %in% df2$names,]
    # name1 name2
    # 1     a     f
    # 2     b     g
    # 4     d     i

or with dplyr

filter(df1, name1 %in% df2$names | name2 %in% df2$names)

# > filter(df1, name1 %in% df2$names | name2 %in% df2$names)
# name1 name2
# 1     a     f
# 2     b     g
# 3     d     i

不确定这是您要寻找的吗？

Answer 3

这就是我最终不确定是否正确的内容。但是在SmitM和TinglTanglBob的代码的帮助下，我想到了：

#This looks for the id number on the steamid column and returns a new variable tf1 with a logical value of T or F. The same goes for the friendid column returning results to tf2
df1$tf1<-df1$steamid %in% df2$V1
df1$tf2<-df1$friendid %in% df2$V1

#The I do two subsets, first a subset of df1 where tf1= TRUE and then a second subest out of that one where tf2= TRUE
df3<-subset(df1,subset = tf1 %in% 'TRUE' & tf2 %in% 'TRUE')
df4<-subset(df3,subset=tf2 %in% 'TRUE')

可悲的是，我结束时的数据比我想象的要少得多。至少如果我做对了。

如果不删除r中的行，则检查df 2中df1中是否存在值

3 个答案: