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How to count occurrences of values of a list in a column of a different dataframe?
How do I check if a list is empty?What is the difference between Python's list methods append and extend?How do I sort a dictionary by value?How to make a flat list out of list of listsHow can I count the occurrences of a list item?How to clone or copy a list?How do I list all files of a directory?Delete column from pandas DataFrame“Large data” work flows using pandasSelect rows from a DataFrame based on values in a column in pandas
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I need to count the number of times the values of a list appear in a column of a different DataFrame.
I've tried using df['Column'].value_counts()
.
However,if a value is in the list but not in the column, it won't show up in the result.
df = pd.DataFrame('Column': [HIGH, HIGH, HIGH, LOW, LOW, LOW, LOW])
list = ['HIGH', 'MEDIUM', 'LOW']
I expect the output to be:
HIGH 3
MEDIUM 0
LOW 4
But using .value_counts()
I get:
HIGH 3
LOW 4
Can someone please point out how I can achieve this? Is it possible to write a piece of code so that my output will always show counts for HIGH, MEDIUM and LOW, regardless of how the data is distributed? (The next time I load the data, it might be possible that there are no HIGH values, instead of MEDIUM)
python pandas group-by jupyter-notebook
add a comment |
I need to count the number of times the values of a list appear in a column of a different DataFrame.
I've tried using df['Column'].value_counts()
.
However,if a value is in the list but not in the column, it won't show up in the result.
df = pd.DataFrame('Column': [HIGH, HIGH, HIGH, LOW, LOW, LOW, LOW])
list = ['HIGH', 'MEDIUM', 'LOW']
I expect the output to be:
HIGH 3
MEDIUM 0
LOW 4
But using .value_counts()
I get:
HIGH 3
LOW 4
Can someone please point out how I can achieve this? Is it possible to write a piece of code so that my output will always show counts for HIGH, MEDIUM and LOW, regardless of how the data is distributed? (The next time I load the data, it might be possible that there are no HIGH values, instead of MEDIUM)
python pandas group-by jupyter-notebook
add a comment |
I need to count the number of times the values of a list appear in a column of a different DataFrame.
I've tried using df['Column'].value_counts()
.
However,if a value is in the list but not in the column, it won't show up in the result.
df = pd.DataFrame('Column': [HIGH, HIGH, HIGH, LOW, LOW, LOW, LOW])
list = ['HIGH', 'MEDIUM', 'LOW']
I expect the output to be:
HIGH 3
MEDIUM 0
LOW 4
But using .value_counts()
I get:
HIGH 3
LOW 4
Can someone please point out how I can achieve this? Is it possible to write a piece of code so that my output will always show counts for HIGH, MEDIUM and LOW, regardless of how the data is distributed? (The next time I load the data, it might be possible that there are no HIGH values, instead of MEDIUM)
python pandas group-by jupyter-notebook
I need to count the number of times the values of a list appear in a column of a different DataFrame.
I've tried using df['Column'].value_counts()
.
However,if a value is in the list but not in the column, it won't show up in the result.
df = pd.DataFrame('Column': [HIGH, HIGH, HIGH, LOW, LOW, LOW, LOW])
list = ['HIGH', 'MEDIUM', 'LOW']
I expect the output to be:
HIGH 3
MEDIUM 0
LOW 4
But using .value_counts()
I get:
HIGH 3
LOW 4
Can someone please point out how I can achieve this? Is it possible to write a piece of code so that my output will always show counts for HIGH, MEDIUM and LOW, regardless of how the data is distributed? (The next time I load the data, it might be possible that there are no HIGH values, instead of MEDIUM)
python pandas group-by jupyter-notebook
python pandas group-by jupyter-notebook
edited Mar 24 at 3:31
ALollz
18.6k51840
18.6k51840
asked Mar 24 at 2:45
pd123pd123
183
183
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
One quick fix reindex
df.Column.value_counts().reindex(list,fill_value=0)
HIGH 3
MEDIUM 0
LOW 4
Name: Column, dtype: int64
Another way pd.Categorical
pd.Categorical(df.Column,list).value_counts()
HIGH 3
MEDIUM 0
LOW 4
dtype: int64
add a comment |
Try below, it creates an empty series first and then combines it with value_counts series:
pd.Series(0, index=list).combine(df.value_counts(), max)
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
One quick fix reindex
df.Column.value_counts().reindex(list,fill_value=0)
HIGH 3
MEDIUM 0
LOW 4
Name: Column, dtype: int64
Another way pd.Categorical
pd.Categorical(df.Column,list).value_counts()
HIGH 3
MEDIUM 0
LOW 4
dtype: int64
add a comment |
One quick fix reindex
df.Column.value_counts().reindex(list,fill_value=0)
HIGH 3
MEDIUM 0
LOW 4
Name: Column, dtype: int64
Another way pd.Categorical
pd.Categorical(df.Column,list).value_counts()
HIGH 3
MEDIUM 0
LOW 4
dtype: int64
add a comment |
One quick fix reindex
df.Column.value_counts().reindex(list,fill_value=0)
HIGH 3
MEDIUM 0
LOW 4
Name: Column, dtype: int64
Another way pd.Categorical
pd.Categorical(df.Column,list).value_counts()
HIGH 3
MEDIUM 0
LOW 4
dtype: int64
One quick fix reindex
df.Column.value_counts().reindex(list,fill_value=0)
HIGH 3
MEDIUM 0
LOW 4
Name: Column, dtype: int64
Another way pd.Categorical
pd.Categorical(df.Column,list).value_counts()
HIGH 3
MEDIUM 0
LOW 4
dtype: int64
answered Mar 24 at 2:48
WeNYoBenWeNYoBen
136k84574
136k84574
add a comment |
add a comment |
Try below, it creates an empty series first and then combines it with value_counts series:
pd.Series(0, index=list).combine(df.value_counts(), max)
add a comment |
Try below, it creates an empty series first and then combines it with value_counts series:
pd.Series(0, index=list).combine(df.value_counts(), max)
add a comment |
Try below, it creates an empty series first and then combines it with value_counts series:
pd.Series(0, index=list).combine(df.value_counts(), max)
Try below, it creates an empty series first and then combines it with value_counts series:
pd.Series(0, index=list).combine(df.value_counts(), max)
answered Mar 24 at 7:34
hacker315hacker315
1,267715
1,267715
add a comment |
add a comment |
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