Replicate each row of data.frame and specify the number of replications for each row?
Update
Upon revisiting this question, I have a feeling that @Codoremifa was correct in their assumption that your "frequency" column might be a factor.
Here's an example if that were the case. It won't match your actual data since I don't know what other levels are in your dataset.
mydf$F2 <- factor(as.character(mydf$frequency))
## expandRows(mydf, "F2")
mydf[rep(rownames(mydf), mydf$F2), ]
# a b frequency F2
# 1 5 3 2 2
# 1.1 5 3 2 2
# 1.2 5 3 2 2
# 2 5 7 1 1
# 3 9 1 40 40
# 3.1 9 1 40 40
# 3.2 9 1 40 40
# 3.3 9 1 40 40
# 4 12 4 5 5
# 4.1 12 4 5 5
# 4.2 12 4 5 5
# 4.3 12 4 5 5
# 4.4 12 4 5 5
# 5 12 5 13 13
# 5.1 12 5 13 13
Hmmm. That doesn't look like 61 rows to me. Why not? Because rep uses the numeric values underlying the factor, which is quite different in this case from the displayed value:
as.numeric(mydf$F2)
# [1] 3 1 4 5 2
To properly convert it, you would need:
as.numeric(as.character(mydf$F2))
# [1] 2 1 40 5 13
Original answer
A while ago I wrote a function that is a bit more of a generalization of @Simono101's answer. The function looks like this:
expandRows <- function(dataset, count, count.is.col = TRUE) {
if (!isTRUE(count.is.col)) {
if (length(count) == 1) {
dataset[rep(rownames(dataset), each = count), ]
} else {
if (length(count) != nrow(dataset)) {
stop("Expand vector does not match number of rows in data.frame")
}
dataset[rep(rownames(dataset), count), ]
}
} else {
dataset[rep(rownames(dataset), dataset[[count]]),
setdiff(names(dataset), names(dataset[count]))]
}
}
For your purposes, you could just use expandRows(mydf, "frequency")
head(expandRows(mydf, "frequency"))
# a b
# 1 5 3
# 1.1 5 3
# 2 5 7
# 3 9 1
# 3.1 9 1
# 3.2 9 1
Other options are to repeat each row the same number of times:
expandRows(mydf, 2, count.is.col=FALSE)
# a b frequency
# 1 5 3 2
# 1.1 5 3 2
# 2 5 7 1
# 2.1 5 7 1
# 3 9 1 40
# 3.1 9 1 40
# 4 12 4 5
# 4.1 12 4 5
# 5 12 5 13
# 5.1 12 5 13
Or to specify a vector of how many times to repeat each row.
expandRows(mydf, c(1, 2, 1, 0, 2), count.is.col=FALSE)
# a b frequency
# 1 5 3 2
# 2 5 7 1
# 2.1 5 7 1
# 3 9 1 40
# 5 12 5 13
# 5.1 12 5 13
Note the required count.is.col = FALSE argument in those last two options.
Repeat each row of data.frame the number of times specified in a column
Here's one solution:
df.expanded <- df[rep(row.names(df), df$freq), 1:2]
Result:
var1 var2
1 a d
2 b e
2.1 b e
3 c f
3.1 c f
3.2 c f
How can I replicate rows in Pandas?
Use np.repeat:
Version 1:
Try using np.repeat:
newdf = pd.DataFrame(np.repeat(df.values, 3, axis=0))
newdf.columns = df.columns
print(newdf)
The above code will output:
Person ID ZipCode Gender
0 12345 882 38182 Female
1 12345 882 38182 Female
2 12345 882 38182 Female
3 32917 271 88172 Male
4 32917 271 88172 Male
5 32917 271 88172 Male
6 18273 552 90291 Female
7 18273 552 90291 Female
8 18273 552 90291 Female
np.repeat repeats the values of df, 3 times.
Then we add the columns with assigning new_df.columns = df.columns.
Version 2:
You could also assign the column names in the first line, like below:
newdf = pd.DataFrame(np.repeat(df.values, 3, axis=0), columns=df.columns)
print(newdf)
The above code will also output:
Person ID ZipCode Gender
0 12345 882 38182 Female
1 12345 882 38182 Female
2 12345 882 38182 Female
3 32917 271 88172 Male
4 32917 271 88172 Male
5 32917 271 88172 Male
6 18273 552 90291 Female
7 18273 552 90291 Female
8 18273 552 90291 Female
Repeat rows of a data.frame
df <- data.frame(a = 1:2, b = letters[1:2])
df[rep(seq_len(nrow(df)), each = 2), ]
Repeat Rows in Data Frame n Times
Use a combination of pd.DataFrame.loc and pd.Index.repeat
test.loc[test.index.repeat(test.times)]
id times
0 a 2
0 a 2
1 b 3
1 b 3
1 b 3
2 c 1
3 d 5
3 d 5
3 d 5
3 d 5
3 d 5
To mimic your exact output, use reset_index
test.loc[test.index.repeat(test.times)].reset_index(drop=True)
id times
0 a 2
1 a 2
2 b 3
3 b 3
4 b 3
5 c 1
6 d 5
7 d 5
8 d 5
9 d 5
10 d 5
Pandas - replicate rows with new column value from a list for each replication
Here is a way using the keys paramater of pd.concat():
(pd.concat([df]*len(New_Cost_List),
keys = New_Cost_List,
names = ['New_Cost',None])
.reset_index(level=0))
Output:
New_Cost State Cost
0 1 A 2
1 1 B 9
2 1 C 8
3 1 D 4
0 5 A 2
1 5 B 9
2 5 C 8
3 5 D 4
0 10 A 2
1 10 B 9
2 10 C 8
3 10 D 4
Repeat rows in a data frame AND add an increment field
You can use uncount from tidyr
library(dplyr)
library(tidyr)
df %>%
uncount(weights = freq, .id = "n", .remove = F) %>%
mutate(value = freq + n - 1)
data startValue freq n value
1 a 3.4 3 1 3
2 a 3.4 3 2 4
3 a 3.4 3 3 5
4 b 2.1 2 1 2
5 b 2.1 2 2 3
6 c 6.3 1 1 1
Python Pandas replicate rows in dataframe
You can put df_try inside a list and then do what you have in mind:
>>> df.append([df_try]*5,ignore_index=True)
Store Dept Date Weekly_Sales IsHoliday
0 1 1 2010-02-05 24924.50 False
1 1 1 2010-02-12 46039.49 True
2 1 1 2010-02-19 41595.55 False
3 1 1 2010-02-26 19403.54 False
4 1 1 2010-03-05 21827.90 False
5 1 1 2010-03-12 21043.39 False
6 1 1 2010-03-19 22136.64 False
7 1 1 2010-03-26 26229.21 False
8 1 1 2010-04-02 57258.43 False
9 1 1 2010-02-12 46039.49 True
10 1 1 2010-02-12 46039.49 True
11 1 1 2010-02-12 46039.49 True
12 1 1 2010-02-12 46039.49 True
13 1 1 2010-02-12 46039.49 True
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