I am new to Pandas and would like to learn how to change alternate column names -
I have a dataframe which looks like this -
I would like to change the column names to be like this -
Any suggestion which apply a for loop would be helpful as I have 57 column names to change in the desired pattern.
A little verbose, but it works:
import pandas as pd
df = pd.DataFrame(columns = ['Time', 'Jolt1', 'Jolt2', 'Time', 'Jolt1', 'Jolt2', 'Time', 'Jolt1', 'Jolt2'])
c = 1
length = int(len(df.columns)/3)
my_list = []
for i in range(length):
my_list.append(['Time', 'Jolt'+str(c), 'Jolt'+str(c + 1)])
c = c + 1
df.columns = sum(my_list, [])
Related
I am relatively new with python - and I am struggling to do the following:
I have a set of different data frames, with sequential naming (df_i), which I want to access in a for loop based on their name (with an string), how can I do that? e.g.
df_1 = pd.read_csv('...')
df_2 = pd.read_csv('...')
df_3 = pd.read_csv('...')
....
n_df = 3
for i in range(len(n_df)):
df_namestr= 'df_' + str(i+1)
# ---------------------
df_temp = df_namestr
# ---------------------
# Operate with df_temp. For i+1= 1, df_temp should be df_1
Kind regards,
DF
You can try something like that:
for n in range(1, n_df+1):
df_namestr = f"df_{n}"
df_tmp = locals().get(df_namestr)
if not isinstance(df_tmp, pd.DataFrame):
continue
print(df_namestr)
print(df_tmp)
Refer to the documentation of locals() to know more.
Would it be better to approach the accessing of multiple dataframes by reading them into a list?
You could put all the csv files required in a subfolder and read them all in. Then they are in a list and you can access each one as an item in that list.
Example:
import pandas as pd
import glob
path = r'/Users/myUsername/Documents/subFolder'
csv_files = glob.glob(path + "/*.csv")
dfs = []
for filename in csv_files:
df = pd.read_csv(filename)
dfs.append(df)
print(len(dfs))
print(dfs[1].head())
I can do the following if I want to extract rows whose column "A" contains the substring "hello".
df[df['A'].str.contains("hello")]
How can I select rows whose column is the substring for another word? e.g.
df["hello".contains(df['A'].str)]
Here's an example dataframe
df = pd.DataFrame.from_dict({"A":["hel"]})
df["hello".contains(df['A'].str)]
IIUC, you could apply str.find:
import pandas as pd
df = pd.DataFrame(['hell', 'world', 'hello'], columns=['A'])
res = df[df['A'].apply("hello".find).ne(-1)]
print(res)
Output
A
0 hell
2 hello
As an alternative use __contains__
res = df[df['A'].apply("hello".__contains__)]
print(res)
Output
A
0 hell
2 hello
Or simply:
res = df[df['A'].apply(lambda x: x in "hello")]
print(res)
df.groupby('columns').apply(''.join()), join all the cells to a string.
This is for a junior dataprocessor. In the past, I've tried many ways.
import pandas as pd
data = {'key':['a','b','c','a','b','c','a'], 'profit':
[12,3,4,5,6,7,9],'income':['j','d','d','g','d','t','d']}
df = pd.DataFrame(data)
df = df.set_index(‘key’)
#df2 is expected result
data2 = {'a':['12j5g9d'],'b':['3d6d'],'c':['4d7t']}
df2 = pd.DataFrame(data2)
df2 = df2.set_index(‘key’)
Here's a simple solution, where we first translate the integers to strings and then concatenate profit and income, then finally we concatenate all strings under the same key:
data = {'key':['a','b','c','a','b','c','a'], 'profit':
[12,3,4,5,6,7,9],'income':['j','d','d','g','d','t','d']}
df = pd.DataFrame(data)
df['profit_income'] = df['profit'].apply(str) + df['income']
res = df.groupby('key')['profit_income'].agg(''.join)
print(res)
output:
key
a 12j5g9d
b 3d6d
c 4d7t
Name: profit_income, dtype: object
This question can be solved couple different ways:
First add an extra column by concatenating the profit and income columns.
import pandas as pd
data = {'key':['a','b','c','a','b','c','a'], 'profit':
[12,3,4,5,6,7,9],'income':['j','d','d','g','d','t','d']}
df = pd.DataFrame(data)
df = df.set_index('key')
df['profinc']=df['profit'].astype(str)+df['income']
1) Using sum
df2=df.groupby('key').profinc.sum()
2) Using apply and join
df2=df.groupby('key').profinc.apply(''.join)
Results from both of the above would be the same:
key
a 12j5g9d
b 3d6d
c 4d7t
I have a dataframe with column names, and I want to find the one that contains a certain string, but does not exactly match it. I'm searching for 'spike' in column names like 'spike-2', 'hey spike', 'spiked-in' (the 'spike' part is always continuous).
I want the column name to be returned as a string or a variable, so I access the column later with df['name'] or df[name] as normal. I've tried to find ways to do this, to no avail. Any tips?
Just iterate over DataFrame.columns, now this is an example in which you will end up with a list of column names that match:
import pandas as pd
data = {'spike-2': [1,2,3], 'hey spke': [4,5,6], 'spiked-in': [7,8,9], 'no': [10,11,12]}
df = pd.DataFrame(data)
spike_cols = [col for col in df.columns if 'spike' in col]
print(list(df.columns))
print(spike_cols)
Output:
['hey spke', 'no', 'spike-2', 'spiked-in']
['spike-2', 'spiked-in']
Explanation:
df.columns returns a list of column names
[col for col in df.columns if 'spike' in col] iterates over the list df.columns with the variable col and adds it to the resulting list if col contains 'spike'. This syntax is list comprehension.
If you only want the resulting data set with the columns that match you can do this:
df2 = df.filter(regex='spike')
print(df2)
Output:
spike-2 spiked-in
0 1 7
1 2 8
2 3 9
This answer uses the DataFrame.filter method to do this without list comprehension:
import pandas as pd
data = {'spike-2': [1,2,3], 'hey spke': [4,5,6]}
df = pd.DataFrame(data)
print(df.filter(like='spike').columns)
Will output just 'spike-2'. You can also use regex, as some people suggested in comments above:
print(df.filter(regex='spike|spke').columns)
Will output both columns: ['spike-2', 'hey spke']
You can also use df.columns[df.columns.str.contains(pat = 'spike')]
data = {'spike-2': [1,2,3], 'hey spke': [4,5,6], 'spiked-in': [7,8,9], 'no': [10,11,12]}
df = pd.DataFrame(data)
colNames = df.columns[df.columns.str.contains(pat = 'spike')]
print(colNames)
This will output the column names: 'spike-2', 'spiked-in'
More about pandas.Series.str.contains.
# select columns containing 'spike'
df.filter(like='spike', axis=1)
You can also select by name, regular expression. Refer to: pandas.DataFrame.filter
df.loc[:,df.columns.str.contains("spike")]
Another solution that returns a subset of the df with the desired columns:
df[df.columns[df.columns.str.contains("spike|spke")]]
You also can use this code:
spike_cols =[x for x in df.columns[df.columns.str.contains('spike')]]
Getting name and subsetting based on Start, Contains, and Ends:
# from: https://stackoverflow.com/questions/21285380/find-column-whose-name-contains-a-specific-string
# from: https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.str.contains.html
# from: https://cmdlinetips.com/2019/04/how-to-select-columns-using-prefix-suffix-of-column-names-in-pandas/
# from: https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.filter.html
import pandas as pd
data = {'spike_starts': [1,2,3], 'ends_spike_starts': [4,5,6], 'ends_spike': [7,8,9], 'not': [10,11,12]}
df = pd.DataFrame(data)
print("\n")
print("----------------------------------------")
colNames_contains = df.columns[df.columns.str.contains(pat = 'spike')].tolist()
print("Contains")
print(colNames_contains)
print("\n")
print("----------------------------------------")
colNames_starts = df.columns[df.columns.str.contains(pat = '^spike')].tolist()
print("Starts")
print(colNames_starts)
print("\n")
print("----------------------------------------")
colNames_ends = df.columns[df.columns.str.contains(pat = 'spike$')].tolist()
print("Ends")
print(colNames_ends)
print("\n")
print("----------------------------------------")
df_subset_start = df.filter(regex='^spike',axis=1)
print("Starts")
print(df_subset_start)
print("\n")
print("----------------------------------------")
df_subset_contains = df.filter(regex='spike',axis=1)
print("Contains")
print(df_subset_contains)
print("\n")
print("----------------------------------------")
df_subset_ends = df.filter(regex='spike$',axis=1)
print("Ends")
print(df_subset_ends)
In the below code, the dataframe df5 is not getting populated. I am just assigning the values to dataframe's columns and I have specified the column beforehand. When I print the dataframe, it returns an empty dataframe. Not sure whether I am missing something.
Any help would be appreciated.
import math
import pandas as pd
columns = ['ClosestLat','ClosestLong']
df5 = pd.DataFrame(columns=columns)
def distance(pt1, pt2):
return math.sqrt((pt1[0] - pt2[0])**2 + (pt1[1] - pt2[1])**2)
for pt1 in df1:
closestPoints = [pt1, df2[0]]
for pt2 in df2:
if distance(pt1, pt2) < distance(closestPoints[0], closestPoints[1]):
closestPoints = [pt1, pt2]
df5['ClosestLat'] = closestPoints[1][0]
df5['ClosestLat'] = closestPoints[1][0]
df5['ClosestLong'] = closestPoints[1][1]
print ("Point: " + str(closestPoints[0]) + " is closest to " + str(closestPoints[1]))
From the look of your code, you're trying to populate df5 with a list of latitudes and longitudes. However, you're making a couple mistakes.
The columns of pandas dataframes are Series, and hold some type of sequential data. So df5['ClosestLat'] = closestPoints[1][0] attempts to assign the entire column a single numerical value, and results in an empty column.
Even if the dataframe wasn't ignoring your attempts to assign a real number to the column, you would lose data because you are overwriting the column with each loop.
The Solution: Build a list of lats and longs, then insert into the dataframe.
import math
import pandas as pd
columns = ['ClosestLat','ClosestLong']
df5 = pd.DataFrame(columns=columns)
def distance(pt1, pt2):
return math.sqrt((pt1[0] - pt2[0])**2 + (pt1[1] - pt2[1])**2)
lats, lngs = [], []
for pt1 in df1:
closestPoints = [pt1, df2[0]]
for pt2 in df2:
if distance(pt1, pt2) < distance(closestPoints[0], closestPoints[1]):
closestPoints = [pt1, pt2]
lats.append(closestPoints[1][0])
lngs.append(closestPoints[1][1])
df['ClosestLat'] = pd.Series(lats)
df['ClosestLong'] = pd.Series(lngs)