I'm trying to add two dataframes using concat with axis = 0, so the columns stay the same but the index increases. One of the dataframes contains a specific columns with a serial number (going from one upwards - but not necessarily in sequence eg. 1,2,3,4,5, etc.)
import pandas as pd
import numpy as np
a = pd.DataFrame(data = {'Name': ['A', 'B','C'],
'Serial Number': [1, 2,5]} )
b = pd.DataFrame(data = {'Name': ['D','E','F'],
'Serial Number': [np.nan,np.nan,np.nan]})
c = pd.concat([a,b],axis=0).reset_index()
I would like to have column 'Serial Number' in dataframe C to start from 5+1 the next one 6+1.
I've tried a variety of things eg:
c.loc[c['B'].isna(), 'B'] = c['B'].shift(1)+1
But it doesn't seem to work.
Desired output:
| Name | Serial Number|
-------------------------
1 A | 1
2 B | 2
3 C | 5
4 D | 6
5 E | 7
6 F | 8
One idea is create arange by number od missinng values add maximal value and 1:
a = np.arange(c['Serial Number'].isna().sum()) + c['Serial Number'].max() + 1
c.loc[c['Serial Number'].isna(), 'Serial Number'] = a
print (c)
index Name Serial Number
0 0 A 1.0
1 1 B 2.0
2 2 C 5.0
3 0 D 6.0
4 1 E 7.0
5 2 F 8.0
Related
Based on the simplifed sample dataframe
import pandas as pd
import numpy as np
timestamps = pd.date_range(start='2017-01-01', end='2017-01-5', inclusive='left')
values = np.arange(0,len(timestamps))
df = pd.DataFrame({'A': values ,'B' : values*2},
index = timestamps )
print(df)
A B
2017-01-01 0 0
2017-01-02 1 2
2017-01-03 2 4
2017-01-04 3 6
I want to use a roll-forward window of size 2 with a stride of 1 to create a resulting dataframe like
timestep_1 timestep_2 target
0 A 0 1 2
B 0 2 4
1 A 1 2 3
B 2 4 6
I.e., each window step should create a data item with the two values of A and B in this window and the A and B values immediately to the right of the window as target values.
My first idea was to use pandas
https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.rolling.html
But that seems to only work in combination with aggregate functions such as sum, which is a different use case.
Any ideas on how to implement this rolling-window-based sampling approach?
Here is one way to do it:
window_size = 3
new_df = pd.concat(
[
df.iloc[i : i + window_size, :]
.T.reset_index()
.assign(other_index=i)
.set_index(["other_index", "index"])
.set_axis([f"timestep_{j}" for j in range(1, window_size)] + ["target"], axis=1)
for i in range(df.shape[0] - window_size + 1)
]
)
new_df.index.names = ["", ""]
print(df)
# Output
timestep_1 timestep_2 target
0 A 0 1 2
B 0 2 4
1 A 1 2 3
B 2 4 6
I have the case where I want to sanity check labeled data. I have hundreds of features and want to find points which have the same features but different label. These found cluster of disagreeing labels should then be numbered and put into a new dataframe.
This isn't hard but I am wondering what the most elegant solution for this is.
Here an example:
import pandas as pd
df = pd.DataFrame({
"feature_1" : [0,0,0,4,4,2],
"feature_2" : [0,5,5,1,1,3],
"label" : ["A","A","B","B","D","A"]
})
result_df = pd.DataFrame({
"cluster_index" : [0,0,1,1],
"feature_1" : [0,0,4,4],
"feature_2" : [5,5,1,1],
"label" : ["A","B","B","D"]
})
In order to get the output you want (both de-duplication and cluster_index), you can use a groupby approach:
g = df.groupby(['feature_1', 'feature_2'])['label']
(df.assign(cluster_index=g.ngroup()) # get group name
.loc[g.transform('size').gt(1)] # filter the non-duplicates
# line below only to have a nice cluster_index range (0,1…)
.assign(cluster_index= lambda d: d['cluster_index'].factorize()[0])
)
output:
feature_1 feature_2 label cluster_index
1 0 5 A 0
2 0 5 B 0
3 4 1 B 1
4 4 1 D 1
First get all duplicated values per feature columns and then if necessary remove duplciated by all columns (here in sample data not necessary), last add GroupBy.ngroup for groups indices:
df = df[df.duplicated(['feature_1','feature_2'],keep=False)].drop_duplicates()
df['cluster_index'] = df.groupby(['feature_1', 'feature_2'])['label'].ngroup()
print (df)
feature_1 feature_2 label cluster_index
1 0 5 A 0
2 0 5 B 0
3 4 1 B 1
4 4 1 D 1
Suppose that I have a pandas dataframe like the one below:
import pandas as pd
df = pd.DataFrame({'fk ID': [1,1,2,2],
'value': [3,3,4,5],
'valID': [1,2,1,2]})
The above would give me the following output:
print(df)
fk ID value valID
0 1 3 1
1 1 3 2
2 2 4 1
3 2 5 2
or
|fk ID| value | valId |
| 1 | 3 | 1 |
| 1 | 3 | 2 |
| 2 | 4 | 1 |
| 2 | 5 | 2 |
and I would like to transpose and pivot it in such a way that I get the following table and the same order of column names:
fk ID value valID fkID value valID
| 1 | 3 | 1 | 1 | 3 | 2 |
| 2 | 4 | 1 | 2 | 5 | 2 |
The most straightforward solution I can think of is
df = pd.DataFrame({'fk ID': [1,1,2,2],
'value': [3,3,4,5],
'valID': [1,2,1,2]})
# concatenate the rows (Series) of each 'fk ID' group side by side
def flatten_group(g):
return pd.concat(row for _, row in g.iterrows())
res = df.groupby('fk ID', as_index=False).apply(flatten_group)
However, using Series.iterrows is not ideal, and can be very slow if the size of each group is large.
Furthermore, the above solution doesn't work if the 'fk ID' groups have different sizes. To see that, we can add a third group to the DataFrame
>>> df2 = df.append({'fk ID': 3, 'value':10, 'valID': 4},
ignore_index=True)
>>> df2
fk ID value valID
0 1 3 1
1 1 3 2
2 2 4 1
3 2 5 2
4 3 10 4
>>> df2.groupby('fk ID', as_index=False).apply(flatten_group)
0 fk ID 1
value 3
valID 1
fk ID 1
value 3
valID 2
1 fk ID 2
value 4
valID 1
fk ID 2
value 5
valID 2
2 fk ID 3
value 10
valID 4
dtype: int64
The result is not a DataFrame as one could expect, because pandas can't align the columns of the groups.
To solve this I suggest the following solution. It should work for any group size, and should be faster for large DataFrames.
import numpy as np
def flatten_group(g):
# flatten each group data into a single row
flat_data = g.to_numpy().reshape(1,-1)
return pd.DataFrame(flat_data)
# group the rows by 'fk ID'
groups = df.groupby('fk ID', group_keys=False)
# get the maximum group size
max_group_size = groups.size().max()
# contruct the new columns by repeating the
# original columns 'max_group_size' times
new_cols = np.tile(df.columns, max_group_size)
# aggregate the flattened rows
res = groups.apply(flatten_group).reset_index(drop=True)
# update the columns
res.columns = new_cols
Output:
# df
>>> res
fk ID value valID fk ID value valID
0 1 3 1 1 3 2
1 2 4 1 2 5 2
# df2
>>> res
fk ID value valID fk ID value valID
0 1 3 1 1.0 3.0 2.0
1 2 4 1 2.0 5.0 2.0
2 3 10 4 NaN NaN NaN
You can cast df as a numpy array, reshape it and cast it back to a dataframe, then rename the columns (0..5).
This is working too if values are not numbers but strings.
import pandas as pd
df = pd.DataFrame({'fk ID': [1,1,2,2],
'value': [3,3,4,5],
'valID': [1,2,1,2]})
nrows = 2
array = df.to_numpy().reshape((nrows, -1))
pd.DataFrame(array).rename(mapper=lambda x: df.columns[x % len(df.columns)], axis=1)
If your group sizes are guaranteed to be the same, you could merge your odd and even rows:
import pandas as pd
df = pd.DataFrame({'fk ID': [1,1,2,2],
'value': [3,3,4,5],
'valID': [1,2,1,2]})
df_even = df[df.index%2==0].reset_index(drop=True)
df_odd = df[df.index%2==1].reset_index(drop=True)
df_odd.join(df_even, rsuffix='_2')
Yields
fk ID value valID fk ID_2 value_2 valID_2
0 1 3 2 1 3 1
1 2 5 2 2 4 1
I'd expect this to be pretty performant, and this could be generalized for any number of rows in each group (vs assuming odd/even for two rows per group), but will require that you have the same number of rows per fk ID.
I have 3 data frame:
df1
id,k,a,b,c
1,2,1,5,1
2,3,0,1,0
3,6,1,1,0
4,1,0,5,0
5,1,1,5,0
df2
name,a,b,c
p,4,6,8
q,1,2,3
df3
type,w_ave,vac,yak
n,3,5,6
v,2,1,4
from the multiplication, using pandas and numpy, I want to the output in df1:
id,k,a,b,c,w_ave,vac,yak
1,2,1,5,1,16,15,18
2,3,0,1,0,0,3,6
3,6,1,1,0,5,4,7
4,1,0,5,0,0,11,14
5,1,1,5,0,13,12,15
the conditions are:
The value of the new column will be =
#its not a code
df1["w_ave"][1] = df3["w_ave"]["v"]+ df1["a"][1]*df2["a"]["q"]+df1["b"][1]*df2["b"]["q"]+df1["c"][1]*df2["c"]["q"]
for output["w_ave"][1]= 2 +(1*1)+(5*2)+(1*3)
df3["w_ave"]["v"]=2
df1["a"][1]=1, df2["a"]["q"]=1 ;
df1["b"][1]=5, df2["b"]["q"]=2 ;
df1["c"][1]=1, df2["c"]["q"]=3 ;
Which means:
- a new column will be added in df1, from the name of the column from df3.
- for each row of the df1, the value of a, b, c will be multiplied with the same-named q value from df2. and summed together with the corresponding value of df3.
-the column name of df1 , matched will column name of df2 will be multiplied. The other not matched column will not be multiplied, like df1[k].
- However, if there is any 0 in df1["a"], the corresponding output will be zero.
I am struggling with this. It was tough to explain also. My attempts are very silly. I know this attempt will not work. However, I have added this:
import pandas as pd, numpy as np
data1 = "Sample_data1.csv"
data2 = "Sample_data2.csv"
data3 = "Sample_data3.csv"
folder = '~Sample_data/'
df1 =pd.read_csv(folder + data1)
df2 =pd.read_csv(folder + data2)
df3 =pd.read_csv(folder + data3)
df1= df2 * df1
Ok, so this will in no way resemble your desired output, but vectorizing the formula you provided:
df2=df2.set_index("name")
df3=df3.set_index("type")
df1["w_ave"] = df3.loc["v", "w_ave"]+ df1["a"].mul(df2.loc["q", "a"])+df1["b"].mul(df2.loc["q", "b"])+df1["c"].mul(df2.loc["q", "c"])
Outputs:
id k a b c w_ave
0 1 2 1 5 1 16
1 2 3 0 1 0 4
2 3 6 1 1 0 5
3 4 1 0 5 0 12
4 5 1 1 5 0 13
I have three columns C1,C2,C3 in panda dataframe. My aim is to replace C1_i by C2_j whenever C3_i=C1_j. These are all strings. I was trying where but failed. What is a good way to do this avoiding for loop?
If my data frame is
df=pd.DataFrame({'c1': ['a', 'b', 'c'], 'c2': ['d','e','f'], 'c3': ['c', 'z', 'b']})
Then I want c3 to be replaced by ['f','z','e']
I tried this, which takes very long time.
for i in range(0,len(df)):
for j in range(0,len(df)):
if (df.iloc[i]['c1']==df.iloc[j]['c3']):
df.iloc[j]['c3']=accounts.iloc[i]['c2']
Use map by Series created by set_index:
df['c3'] = df['c3'].map(df.set_index('c1')['c2']).fillna(df['c3'])
Alternative solution with update:
df['c3'].update(df['c3'].map(df.set_index('c1')['c2']))
print (df)
c1 c2 c3
0 a d f
1 b e z
2 c f e
Example data:
dataframe = pd.DataFrame({'a':['10','4','3','40','5'], 'b':['5','4','3','2','1'], 'c':['s','d','f','g','h']})
Output:
a b c
0 10 5 s
1 4 4 d
2 3 3 f
3 40 2 g
4 5 1 h
Code:
def replace(df):
if len(dataframe[dataframe.b==df.a]) != 0:
df['a'] = dataframe[dataframe.b==df.a].c.values[0]
return df
dataframe = dataframe.apply(replace, 1)
Output:
a b c
0 1 5 0
1 2 4 0
2 0 3 0
3 4 2 0
4 5 1 0
Is it what you want?