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Should I use a data.frame or a matrix?
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edited Aug 30 '17 at 0:12
answered Mar 1 '11 at 19:00
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Android微信智能心跳方案前言：在13年11月中旬时，因为基础组件组人手紧张，Leo安排我和春哥去广州轮岗支援。刚到广州的时候，Ray让我和春哥对Line和WhatsApp的心跳...
前言：在13年11月中旬时，因为基础组件组人手紧张，Leo安...
Automatically expanding an R factor into a collection of 1/0 indicator variables for every factor le
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Use the model.matrix function:
model.matrix( ~ Species - 1, data=iris )
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What's the fastest way to merge/join data.frames in R?
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DF1 = data.frame(a = c(1, 1, 2, 2), b = 1:4)
DF2 = data.frame(b = c(1, 2, 3, 3, 4), c = letters[1:5])
merge(DF1, DF2)
b a c
1 1 1 a
2 2 1 b
3 3 2 c
4 3 2 d
5 4 2 e
DF1$c = DF2$c[match(DF1$b, DF2$b)]
DF1$c
[1] a b c e
Levels: a b c d e
> DF1
a b c
1 1 1 a
2 1 2 b
3 2 3 c
4 2 4 e
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How do I replace NA values with zeros in an R dataframe?
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See my comment in @gsk3 answer. A simple example:
> m <- matrix(sample(c(NA, 1:10), 100, replace = TRUE), 10)
> d <- as.data.frame(m)
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
1 4 3 NA 3 7 6 6 10 6 5
2 9 8 9 5 10 NA 2 1 7 ...
How does virtual inheritance solve the “diamond” (multiple inheritance) ambiguity?
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edited Apr 23 '18 at 5:23
StoryTeller - Unslander Monica
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Access lapply index names inside FUN
...uments to the function, so the following works:
x <- list(a=11,b=12,c=13) # Changed to list to address concerns in commments
lapply(seq_along(x), function(y, n, i) { paste(n[[i]], y[[i]]) }, y=x, n=names(x))
Here I use lapply over the indices of x, but also pass in x and the names of x. As you...
Append a dictionary to a dictionary [duplicate]
... For example,
>>> d1 = {1: 1, 2: 2}
>>> d2 = {2: 'ha!', 3: 3}
>>> d1.update(d2)
>>> d1
{1: 1, 2: 'ha!', 3: 3}
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Python: Find in list
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1235
As for your first question: that code is perfectly fine and should work if item equals one of t...