I'm struggling to understand how the parameters for df.groupby works. I have the following code:
df = pd.read_sql(query_cnxn)
codegroup = df.groupby(['CODE'])
I then attempt a for loop as follows:
for code in codegroup:
dfsize = codegroup.size()
dfmax = codegroup['ID'].max()
dfmin = codegroup['ID'].min()
result = ((dfmax-dfmin)-dfsize)
if result == 1:
df2 = df2.append(itn)
else:
df3 = df3.append(itn)
I'm trying to iterate over each unique code. Does the for loop understand that i'm trying to loop through each code based on the above? Thank you in advance.
Pandas groupby returns an iterator that emits the key of the iterating group and group df as a tuple. You can perform your max and min operation on the group as:
In [1]: import pandas as pd
In [2]: df = pd.DataFrame({'a': [0, 0, 0, 1, 1, 1], 'b': [3, 4, 5, 6, 7, 8]})
In [3]: for k, g in df.groupby('a'):
...: print(g['b'].max())
...:
5
8
You can also get the min-max directly as df using agg:
In [4]: df.groupby('a')['b'].agg(['max', 'min'])
Out[4]:
max min
a
0 5 3
1 8 6
Related
I wanted to create a simple script, which counts values in one column, that are higher in another column:
d = {'a': [1, 3], 'b': [0, 2]}
df = pd.DataFrame(data=d, index=[1, 2])
print(df)
a b
1 1 0
2 3 2
My function:
def diff(dataframe):
a_counter=0
b_counter=0
for i in dataframe["a"]:
for ii in dataframe["b"]:
if i>ii:
a_counter+=1
elif ii>i:
b_counter+=1
return a_counter, b_counter
However
diff(df)
returns (3, 1), instead of (2,0). I know the problem is that every single value of one column gets compared to every value of the other column (e.g. 1 gets compared to 0 and 2 of column b). There probably is a special function for my problem, but can you help me fix my script?
I would suggest adding some helper columns in an intuitive way to help compute the sum of each condition a > b and b > a
A working example based on your code :
import numpy as np
import pandas as pd
d = {'a': [1, 3], 'b': [0, 2]}
df = pd.DataFrame(data=d, index=[1, 2])
def diff(dataframe):
dataframe['a>b'] = np.where(dataframe['a']>dataframe['b'], 1, 0)
dataframe['b>a'] = np.where(dataframe['b']>dataframe['a'], 1, 0)
return dataframe['a>b'].sum(), dataframe['b>a'].sum()
print(diff(df))
>>> (2, 0)
Basically what np.where() does, the way I used it, is that it produces 1 if the condition is met and 0 otherwise. You can then add those columns up using a simple sum() function applied on the desired columns.
Update
Maybe you can use:
>>> df['a'].gt(df['b']).sum(), df['b'].gt(df['a']).sum()
(2, 0)
IIUC, to fix your code:
def diff(dataframe):
a_counter=0
b_counter=0
for i in dataframe["a"]:
for ii in dataframe["b"]:
if i>ii:
a_counter+=1
elif ii>i:
b_counter+=1
# Subtract the minimum of counters
m = min(a_counter, b_counter)
return a_counter-m, b_counter-m
Output:
>>> diff(df)
(2, 0)
IIUC, you can use the sign of the difference and count the values:
d = {1: 'a', -1: 'b', 0: 'equal'}
(np.sign(df['a'].sub(df['b']))
.map(d)
.value_counts()
.reindex(list(d.values()), fill_value=0)
)
output:
a 2
b 0
equal 0
dtype: int64
I have the next csv and I need get the values duplicated from DialedNumer column and then the averege Duration of those duplicates.
I already got the duplicates with the next code:
df = pd.read_csv('cdrs.csv')
dnidump = pd.DataFrame(df, columns=['DialedNumber'])
pd.options.display.float_format = '{:.0f}'.format
dupl_dni = dnidump.pivot_table(index=['DialedNumber'], aggfunc='size')
a1 = dupl_dni.to_frame().rename(columns={0:'TimesRepeated'}).sort_values(by=['TimesRepeated'], ascending=False)
b = a1.head(10)
print(b)
Output:
DialedNumber TimesRepeated
50947740194 4
50936564292 2
50931473242 3
I can't figure out how to get the duration avarege of those duplicates, any ideas?
thx
try:
df_mean = df.groupby('DialedNumber').mean()
Use df.groupby('column').mean()
Here is sample code.
Input
df = pd.DataFrame({'A': [1, 1, 1, 2, 2],
'B': [2461, 1023, 9, 5614, 212],
'C': [2, 4, 8, 16, 32]}, columns=['A', 'B', 'C'])
df.groupby('A').mean()
Output
B C
A
1 1164.333333 4.666667
2 2913.000000 24.000000
API reference of pandas.core.groupby.GroupBy.mean
https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.core.groupby.GroupBy.mean.html
I'm trying to write a function that take a pandas Dataframe as argument and at some concatenate this datagframe with another.
for exemple:
def concat(df):
df = pd.concat((df, pd.DataFrame({'E': [1, 1, 1]})), axis=1)
I would like this function to modify in place the input df but I can't find how to achieve this. When I do
...
print(df)
concat(df)
print(df)
The dataframe df is identical before and after the function call
Note: I don't want to do df['E'] = [1, 1, 1] because I don't know how many column will be added to df. So I want to use pd.concat(), if possible...
This will edit the original DataFrame inplace and give the desired output as long as the new data contains the same number of rows as the original, and there are no conflicting column names.
It's the same idea as your df['E'] = [1, 1, 1] suggestion, except it will work for an arbitrary number of columns.
I don't think there is a way to achieve this using pd.concat, as it doesn't have an inplace parameter as some Pandas functions do.
df = pd.DataFrame({'A': [1, 2, 3], 'B': [4, 5, 6]})
df2 = pd.DataFrame({'C': [10, 20, 30], 'D': [40, 50, 60]})
df[df2.columns] = df2
Results (df):
A B C D
0 1 4 10 40
1 2 5 20 50
2 3 6 30 60
I am trying to place multiple lists into a single column of a Pandas df. My list of lists is very long, so I cannot do so manually.
The desired out put would look like this:
list_of_lists = [[1,2,3],[3,4,5],[5,6,7],...]
df = pd.DataFrame(list_of_lists)
>>> df
0
0 [1,2,3]
1 [3,4,5]
2 [5,6,7]
3 ...
Thank you for the assistance.
You can assign it by wrapping it in a Series vector if you're trying to add to an existing df:
In [7]:
import numpy as np
import pandas as pd
df = pd.DataFrame(np.random.randn(5,3), columns=list('abc'))
df
Out[7]:
a b c
0 -1.675422 -0.696623 -1.025674
1 0.032192 0.582190 0.214029
2 -0.134230 0.991172 -0.177654
3 -1.688784 1.275275 0.029581
4 -0.528649 0.858710 -0.244512
In [9]:
df['new_col'] = pd.Series([[1,2,3],[3,4,5],[5,6,7]])
df
Out[9]:
a b c new_col
0 -1.675422 -0.696623 -1.025674 [1, 2, 3]
1 0.032192 0.582190 0.214029 [3, 4, 5]
2 -0.134230 0.991172 -0.177654 [5, 6, 7]
3 -1.688784 1.275275 0.029581 NaN
4 -0.528649 0.858710 -0.244512 NaN
What about
df = pd.DataFrame({0: [[1,2,3],[3,4,5],[5,6,7]]})
The above solutions were helpful but wanted to add a little bit in case they didn't quite do the trick for someone...
pd.Series will not accept a np.ndarray that looks like a list-of-lists, e.g. one-hot labels array([[1, 0, 0], [0, 1, 0], ..., [0, 0, 1]]).
So in this case one can wrap the variable with list():
df['new_col'] = pd.Series(list(one-hot-labels))
I use groupby on a dataframe based on the columns I want and then I have to take the index of each item in its group. By index I mean, if there are 10 items in a group, the index goes from 0 to 9, not the dataframe index.
My code for doing this is below:
import pandas as pd
df = pd.DataFrame({'A': np.random.randint(0, 11, 10 ** 3), 'B': np.random.randint(0, 11, 10 ** 3),
'C': np.random.randint(0, 11, 10 ** 3), 'D': np.random.randint(0, 2, 10 ** 3)})
grouped_by = df.groupby(["A", "B", "C"])
groups = dict(list(grouped_by))
index_dict = {k: v.index.tolist() for k,v in groups.items()}
df["POS"] = df.apply(lambda x: index_dict[(x["A"], x["B"], x["C"])].index(x.name), axis=1)
The dataframe here is just an example.
Is there a way to use the grouped_by to achieve this ?
Here's a solution using cumcount() on a dummy variable to generate a item index for each group. It should be significantly faster too.
In [122]: df['dummy'] = 0
...: df["POS"] = df.groupby(['A','B','C'])['dummy'].cumcount()
...: df = df.drop('dummy', axis=1)
As #unutbu noted, even cleaner just to use:
df["POS"] = df.groupby(['A','B','C']).cumcount()