I have a data frame :
A B
1 2
4 3
5 9
6 7
9 7
I want to check if values in column A are divisible by 2 (check odd even) if they are divisible by two then I want to add 18 to the value in Column B
So far I have been able to check if value in column A is divisible by 2 and extract it.
df = df[df['A'] % 2 == 0]
Thanks
df['A']%2==0 will return boolean series where A is divisible by 2 and then corresponding values of B would be updated
df.loc[df['A']%2==0, 'B'] = df['B'] + 18
df
A B
0 1 2
1 4 21
2 5 9
3 6 25
4 9 7
Lets try:
df['B']=np.where(df['A'] % 2 == 0,df.B.add(18),df.B)
Related
[enter image description here][1]
Index number 72 is missing from original dataframe which is shown in image. I want to cut dataframe like [0:71,:] with condition like when index sequence breaks then dataframe automatically cuts from previous index value.
Compare shifted values of index subtracted by original values if greater like 1 with invert ordering by [::-1] and Series.cummax, last filter in boolean indexing:
df = pd.DataFrame({'a': range(3,13)}).drop(3)
print (df)
a
0 3
1 4
2 5
4 7
5 8
6 9
7 10
8 11
9 12
df = df[df.index.to_series().shift(-1, fill_value=0).sub(df.index).gt(1)[::-1].cummax()]
print (df)
a
0 3
1 4
2 5
i came to this:
df = pd.DataFrame({'col':[1,2,3,4,5,6,7,8,9]}, index=[-1,0,1,2,3,4,5,7,8])
ind = next((i for i in range(len(df)-1) if df.index[i]+1!=df.index[i+1]),len(df))+1
>>> df.iloc[:ind]
'''
col
-1 1
0 2
1 3
2 4
3 5
4 6
5 7
With numpy, get the values that are equal to a normal range starting from the first index, up to the first mismatch (excluded):
df[np.minimum.accumulate(df.index==np.arange(df.index[0], df.index[0]+len(df)))]
Example:
col
-1 1
0 2
1 3
3 4
4 5
output:
col
-1 1
0 2
1 3
:)
I've a dataframe like that (it's an extract of the entire dataframe):
a b
1 1
6 3
7 5
1 7
12 5
12 5
2 5
95 2
44 3
i want to create a new column using NumPy in python based on a multiple where conditions, considering previous conditions. Let me explain with an example:
I want to create column 'C' with value = '1' when:
(a > b) and (a[-1] < b) and (the previous valued value of "c" must be 2)
another condition is 'C' = '2' when:
(a < b) and (the previous valued value of "c" must be 1)
Thanks you!
You can use np.select to return an array drawn from elements in choicelist, depending on conditions.
Use:
df['c'] = '' # --> assign initial value
conditions = [
(df['a'].gt(df['b']) & df['a'].shift().lt(df['b'])) & (df['c'].shift().eq('') | df['c'].shift().eq(2)),
df['a'].lt(df['b']) & (df['c'].shift().eq(1) | df['c'].shift().eq(''))
]
choices = [1, 2]
df['c'] = np.select(conditions, choices, default='')
print(df)
This prints:
a b c
0 1 1
1 6 3 1
2 7 5
3 1 7 2
4 12 5 1
5 12 5
6 2 5 2
7 95 2
8 44 3
I have a table with 40 columns and 1500 rows. I want to find the maximum value among the 30-32nd (3 columns). How can it be done? I want to return the maximum value among these 3 columns and the index of dataframe.
print(Max_kVA_df.iloc[30:33].max())
hi you can refer this example
import pandas as pd
df=pd.DataFrame({'col1':[1,2,3,4,5],
'col2':[4,5,6,7,8],
'col3':[2,3,4,5,7]
})
print(df)
#print(df.iloc[:,0:3].max())# Mention range of the columns which you want, In your case change 0:3 to 30:33, here 33 will be excluded
ser=df.iloc[:,0:3].max()
print(ser.max())
Output
8
Select values by positions and use np.max:
Sample: for maximum by first 5 rows:
np.random.seed(123)
df = pd.DataFrame(np.random.randint(10, size=(10, 3)), columns=list('ABC'))
print (df)
A B C
0 2 2 6
1 1 3 9
2 6 1 0
3 1 9 0
4 0 9 3
print (df.iloc[0:5])
A B C
0 2 2 6
1 1 3 9
2 6 1 0
3 1 9 0
4 0 9 3
print (np.max(df.iloc[0:5].max()))
9
Or use iloc this way:
print(df.iloc[[30, 31], 2].max())
I want to sort a subset of a dataframe (say, between indexes i and j) according to some value. I tried
df2=df.iloc[i:j].sort_values(by=...)
df.iloc[i:j]=df2
No problem with the first line but nothing happens when I run the second one (not even an error). How should I do ? (I tried also the update function but it didn't do either).
I believe need assign to filtered DataFrame with converting to numpy array by values for avoid align indices:
df = pd.DataFrame({'A': [1,2,3,4,3,2,1,4,1,2]})
print (df)
A
0 1
1 2
2 3
3 4
4 3
5 2
6 1
7 4
8 1
9 2
i = 2
j = 7
df.iloc[i:j] = df.iloc[i:j].sort_values(by='A').values
print (df)
A
0 1
1 2
2 1
3 2
4 3
5 3
6 4
7 4
8 1
9 2
I have a large time series df (2.5mil rows) that contain 0 values in a given row, some of which are legitimate. However if there are repeated continuous occurrences of zero values I would like to remove them from my df.
Example:
Col. A contains [1,2,3,0,4,5,0,0,0,1,2,3,0,8,8,0,0,0,0,9] I would like to remove the [0,0,0] and [0,0,0,0] from the middle and leave the remaining 0 to make a new df [1,2,3,0,4,5,1,2,3,0,8,8,9].
The length of zero values before deletion being a parameter that has to be set - in this case > 2.
Is there a clever way to do this in pandas?
It looks like you want to remove the row if it is 0 and either previous or next row in same column is 0. You can use shift to look for previous and next value and compare with current value as below:
result_df = df[~(((df.ColA.shift(-1) == 0) & (df.ColA == 0)) | ((df.ColA.shift(1) == 0) & (df.ColA == 0)))]
print(result_df)
Result:
ColA
0 1
1 2
2 3
3 0
4 4
5 5
9 1
10 2
11 3
12 0
13 8
14 8
19 9
Update for more than 2 consecutive
Following example in link, adding new column to track consecutive occurrence and later checking it to filter:
# https://stackoverflow.com/a/37934721/5916727
df['consecutive'] = df.ColA.groupby((df.ColA != df.ColA.shift()).cumsum()).transform('size')
df[~((df.consecutive>10) & (df.ColA==0))]
We need build a new para meter here, then using drop_duplicates
df['New']=df.A.eq(0).astype(int).diff().ne(0).cumsum()
s=pd.concat([df.loc[df.A.ne(0),:],df.loc[df.A.eq(0),:].drop_duplicates(keep=False)]).sort_index()
s
Out[190]:
A New
0 1 1
1 2 1
2 3 1
3 0 2
4 4 3
5 5 3
9 1 5
10 2 5
11 3 5
12 0 6
13 8 7
14 8 7
19 9 9
Explanation :
#df.A.eq(0) to find the value equal to 0
#diff().ne(0).cumsum() if they are not equal to 0 then we would count them in same group .