Count Number of Occurences For Each Unique Value

Count number of occurences for each unique value

Perhaps table is what you are after?

dummyData = rep(c(1,2, 2, 2), 25)

table(dummyData)
# dummyData
#  1  2 
# 25 75

## or another presentation of the same data
as.data.frame(table(dummyData))
#    dummyData Freq
#  1         1   25
#  2         2   75

Get number of occurrences of each unique value

df %>% select(val) %>% group_by(val) %>% mutate(count = n()) %>% unique()

This first filters out the value of interest, groups by it, then creates a new column all the unique values, and the number of occurrences of each of those values.

Here is a reproducible example showcasing how it works:

id <- c(1,2,3,4,5,6,7,8,9,0)
val <- c(0,1,2,3,1,1,1,0,0,2)
df <- data.frame(id=id,val=val)
df
#>    id val
#> 1   1   0
#> 2   2   1
#> 3   3   2
#> 4   4   3
#> 5   5   1
#> 6   6   1
#> 7   7   1
#> 8   8   0
#> 9   9   0
#> 10  0   2

df %>% select(val) %>% group_by(val) %>% mutate(count = n()) %>% unique()
#> # A tibble: 4 x 2
#> # Groups:   val [4]
#>     val count
#>   <dbl> <int>
#> 1     0     3
#> 2     1     4
#> 3     2     2
#> 4     3     1

^{Created on 2020-06-17 by the reprex package (v0.3.0)}

Count number of occurrences for each unique value

Yes, you can use GROUP BY:

SELECT time,
       activities,
       COUNT(*)
FROM table
GROUP BY time, activities;

how to count occurrence of each unique value in pandas

It seems like you want to compute value counts across all columns. You can flatten it to a series, drop NaNs, and call value_counts. Here's a sample -

df

     a    b
0  1.0  NaN
1  1.0  NaN
2  3.0  3.0
3  NaN  4.0
4  5.0  NaN
5  NaN  4.0
6  NaN  5.0

pd.Series(df.values.ravel()).dropna().value_counts()

5.0    2
4.0    2
3.0    2
1.0    2
dtype: int64

Another method is with np.unique -

u, c = np.unique(pd.Series(df.values.ravel()).dropna().values, return_counts=True)
pd.Series(c, index=u)

1.0    2
3.0    2
4.0    2
5.0    2
dtype: int64

Note that the first method sorts results in descending order of counts, while the latter does not.

Count unique occurrences within data frame

One option could be:

sapply(df, function(x) table(factor(x, levels = unique(unlist(df)))))

  V1 v2 v3
A  1  1  2
B  1  2  0
D  1  0  1
C  0  1  1

How to count unique values a column in R

We can use n_distinct() from dplyr to count the number of unique values for a column in a data frame.

textFile <- "id var1
111   A
109   A
112   A
111   A
108   A"

df <- read.table(text = textFile,header = TRUE)
library(dplyr)
df %>% summarise(count = n_distinct(id))

...and the output:

> df %>% summarise(count = n_distinct(id))
  count
1     4

We can also summarise the counts within one or more by_group() columns.

textFile <- "id var1
111   A
109   A
112   A
111   A
108   A
201   B
202   B
202   B
111   B
112   B
109   B"

df <- read.table(text = textFile,header = TRUE)
df %>%  group_by(var1) %>% summarise(count = n_distinct(id))

...and the output:

`summarise()` ungrouping output (override with `.groups` argument)
# A tibble: 2 x 2
  var1  count
  <chr> <int>
1 A         4
2 B         5

How to get unique values with respective occurrence count from a list in Python?

If your items are grouped (i.e. similar items come together in a bunch), the most efficient method to use is itertools.groupby:

>>> [(g[0], len(list(g[1]))) for g in itertools.groupby(['a', 'a', 'b', 'b', 'b'])]
[('a', 2), ('b', 3)]

How to find out number of unique values in a column along with count of the unique values in a dataframe?

Use pd.value_counts

pd.value_counts(df.Account_Type)

Gold        3
Platinum    1
Name: Account_Type, dtype: int64

Get number of unique as well

s = pd.value_counts(df.Account_Type)
s1 = pd.Series({'nunique': len(s), 'unique values': s.index.tolist()})
s.append(s1)

Gold                            3
Platinum                        1
nunique                         2
unique values    [Gold, Platinum]
dtype: object

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Alternate Approach

df['col1'].value_counts(sort=True)
df['col1'].value_counts(sort=True, normalize=True) -> provides proportion

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