I have a sheet of numbers, separated by spaces into columns. Each column represents a different category, and within each column, each number represents a different value. For example, column number four represents age, and within the column, the number 5 represents an age of 44-55. Obviously, each row is a different person's record. I'd like to use a Python script to search through the the sheet, and find all columns where the sixth column is number "1." After that, I want to know how many times each number in column one appears where the number in column six is equal to "1." The script should output to the user that "While column six equals '1', the value '1' appears 12 times in column one. The value '2' appears 18 times..." etc. I hope I'm being clear here. I just want it to list the numbers, basically. Anyway, I'm new to Python. I've attached my code below. I think I should be using dictionaries, but I'm just not totally sure how. So far, I haven't really come close to figuring this out. I would really appreciate if someone could walk me through the logic that would be behind such code. Thank you so much!
ldata = open("list.data", "r")
income_dist = {}
for line in ldata:
linelist = line.strip().split(" ")
key_income_dist = linelist[6]
if key_income_dist in income_dist:
income_dist[key_income_dist] = 1 + income_dist[key_income_dist]
else:
income_dist[key_income_dist] = 1
ldata.close()
print value_no_occupations
First, indentation is majorly important in Python and the above is bad: the 5 lines following linelist = line.strip().split(" ") need to be indented to be in the loop like they should be.
Next they should be indented further and this line added before them:
if len(linelist)>6 and linelist[6]=="1":
This line skips over short lines (there are some), and tests for what you said you wanted: "where column six equals "1."" This is column [6] where the first number on the line is referenced as [0] (these are "offsets", not "cardinal", or counting, numbers).
You'll probably want to change key_income_dist = linelist[6] to key_income_dist = linelist[0] or [1] to get what you want. Play around if necessary.
Finally, you should say print income_dist at the end to get a look at your results. If you want fancier output, study up on formatting.
This is actually easier than it seems! The key is collections.Counter
from collections import Counter
ldata = open("list.data")
rows = [tuple(row.split()) for row in ldata if row.split()[5]==1]
# warning this will break if some rows are shorter than 6 columns
first_col = Counter(item[0] for item in rows)
If you want the distribution of every column (not just the first) do:
distribution = {column: Counter(item[column] for item in rows) for column in range(len(rows[0]))}
# warning this will break if all rows are not the same size!
Considering that the data file has ~9000 rows of data, if you don't want to keep the original data, you can combine step 1 & 2 to make the program use less memory and a little faster.
ldata = open("list.data", "r")
# read in all the rows, note that the list values are strings instead of integers
# keep only the rows with 6th column = '1'
only1 = []
for line in ldata:
if line.strip() == '': # ignor blank lines
continue
row = tuple(line.strip().split(" "))
if row[5] == '1':
only1.append(row)
ldata.close()
# tally the statistics
income_dist = {}
for row in only1:
if row[0] in income_dist:
income_dist[row[0]] += 1
else:
income_dist[row[0]] = 1
# print result
print "While column six equals '1',"
for num in sorted(income_dist):
print "the value %s appears %d times in column one." % (num, income_dist[num])
Sample Test Data in list.data:
9 2 1 5 4 5 5 3 3 0 1 1 7 NA
9 1 1 5 5 5 5 3 5 2 1 1 7 1
9 2 1 3 5 1 5 2 3 1 2 3 7 1
1 2 5 1 2 6 5 1 4 2 3 1 7 1
1 2 5 1 2 6 3 1 4 2 3 1 7 1
8 1 1 6 4 8 5 3 2 0 1 1 7 1
1 1 5 2 3 9 4 1 3 1 2 3 7 1
6 1 3 3 4 1 5 1 1 0 2 3 7 1
2 1 1 6 3 8 5 3 3 0 2 3 7 1
4 1 1 7 4 8 4 3 2 0 2 3 7 1
1 1 5 2 4 1 5 1 1 0 2 3 7 1
4 2 2 2 3 2 5 1 2 0 1 1 5 1
8 2 1 3 6 6 2 2 4 2 1 1 7 1
7 2 1 5 3 5 5 3 4 0 2 1 7 1
1 1 5 2 3 9 4 1 3 1 2 3 7 1
6 1 3 3 4 1 5 1 1 0 2 3 7 1
2 1 1 6 3 8 5 3 3 0 2 3 7 1
4 1 1 7 4 8 4 3 2 0 2 3 7 1
1 1 5 2 4 9 5 1 1 0 2 3 7 1
4 2 2 2 3 2 5 1 2 0 1 1 5 1
Following your original program logic, I come up with this version:
ldata = open("list.data", "r")
# read in all the rows, note that the list values are strings instead of integers
linelist = []
for line in ldata:
linelist.append(tuple(line.strip().split(" ")))
ldata.close()
# keep only the rows with 6th column = '1'
only1 = []
for row in linelist:
if row[5] == '1':
only1.append(row)
# tally the statistics
income_dist = {}
for row in only1:
if row[0] in income_dist:
income_dist[row[0]] += 1
else:
income_dist[row[0]] = 1
# print result
print "While column six equals '1',"
for num in sorted(income_dist):
print "the value %s appears %d times in column one." % (num, income_dist[num])
Related
I would like to create the following pattern:
1
2 2
3 3 3
4 4 4 4
5 5 5 5 5
Here is my attempt:
def print_numbers(number):
for i in range(number):
print(i * i)
Now clearly this will return the product of the number with itself. I would like to specify this function to print out the number that many times. So something like print(i) * i which is not correct syntax of course. How would I about doing this?
As mentioned by #Barmar you can convert the number into strings and multiply with same number.
#Ex str(4)*3 = 444 #It will print 3times 4.
def print_numbers(number):
for i in range(number+1):
print((str(i)+" ")*i) #Converted number into string and multiply with the same number.
print_numbers(5)
Output:
1
2 2
3 3 3
4 4 4 4
5 5 5 5 5
If you wanted a line space b/w each number you can print like: print((str(i)+" ")*i+"\n") This will give.
1
2 2
3 3 3
4 4 4 4
5 5 5 5 5
I am working on a small task in which I have to find the distance between two nodes. Each node has X and Y coordinates which can be seen below.
node_number X_coordinate Y_coordinate
0 0 1 0
1 1 1 1
2 2 1 2
3 3 1 3
4 4 0 3
5 5 0 4
6 6 1 4
7 7 2 4
8 8 3 4
9 9 4 4
10 10 4 3
11 11 3 3
12 12 2 3
13 13 2 2
14 14 2 1
15 15 2 0
For the purpose I mentioned above, I wrote below code,
X1_coordinate = df['X_coordinate'].tolist()
Y1_coordinate = df['Y_coordinate'].tolist()
node_number1 = df['node_number'].tolist()
nodal_dist = []
i = 0
for i in range(len(node_number1)):
dist = math.sqrt((X1_coordinate[i+1] - X1_coordinate[i])**2 + (Y1_coordinate[i+1] - Y1_coordinate[i])**2)
nodal_dist.append(dist)
I got the error
list index out of range
Kindly let me know what I am doing wrong and what should I change to get the answer.
Indexing starts at zero, so the last element in the list has an index that is one less than the number of elements in that list. But the len() function gives you the number of elements in the list (in other words, it starts counting at 1), so you want the range of your loop to be len(node_number1) - 1 to avoid an -off-by-one error.
The problems should been in this line
dist = math.sqrt((X1_coordinate[i+1] - X1_coordinate[i])**2 + (Y1_coordinate[i+1] - Y1_coordinate[i])**2)
the X1_coordinate[i+1] and the ] Y1_coordinate[i+1]] go out of range on the last number call.
Assume I have the following pandas data frame:
my_class value
0 1 1
1 1 2
2 1 3
3 2 4
4 2 5
5 2 6
6 2 7
7 2 8
8 2 9
9 3 10
10 3 11
11 3 12
I want to identify the indices of "my_class" where the class changes and remove n rows after and before this index. The output of this example (with n=2) should look like:
my_class value
0 1 1
5 2 6
6 2 7
11 3 12
My approach:
# where class changes happen
s = df['my_class'].ne(df['my_class'].shift(-1).fillna(df['my_class']))
# mask with `bfill` and `ffill`
df[~(s.where(s).bfill(limit=1).ffill(limit=2).eq(1))]
Output:
my_class value
0 1 1
5 2 6
6 2 7
11 3 12
One of possible solutions is to:
Make use of the fact that the index contains consecutive integers.
Find index values where class changes.
For each such index generate a sequence of indices from n-2
to n+1 and concatenate them.
Retrieve rows with indices not in this list.
The code to do it is:
ind = df[df['my_class'].diff().fillna(0, downcast='infer') == 1].index
df[~df.index.isin([item for sublist in
[ range(i-2, i+2) for i in ind ] for item in sublist])]
my_class = np.array([1] * 3 + [2] * 6 + [3] * 3)
cols = np.c_[my_class, np.arange(len(my_class)) + 1]
df = pd.DataFrame(cols, columns=['my_class', 'value'])
df['diff'] = df['my_class'].diff().fillna(0)
idx2drop = []
for i in df[df['diff'] == 1].index:
idx2drop += range(i - 2, i + 2)
print(df.drop(idx_drop)[['my_class', 'value']])
Output:
my_class value
0 1 1
5 2 6
6 2 7
11 3 12
I'm trying to parse a logfile of our manufacturing process. Most of the time the process is run automatically but occasionally, the engineer needs to switch into manual mode to make some changes and then switches back to automatic control by the reactor software. When set to manual mode the logfile records the step as being "MAN.OP." instead of a number. Below is a representative example.
steps = [1,2,2,'MAN.OP.','MAN.OP.',2,2,3,3,'MAN.OP.','MAN.OP.',4,4]
ser_orig = pd.Series(steps)
which results in
0 1
1 2
2 2
3 MAN.OP.
4 MAN.OP.
5 2
6 2
7 3
8 3
9 MAN.OP.
10 MAN.OP.
11 4
12 4
dtype: object
I need to detect the 'MAN.OP.' and make them distinct from each other. In this example, the two regions with values == 2 should be one region after detecting the manual mode section like this:
0 1
1 2
2 2
3 Manual_Mode_0
4 Manual_Mode_0
5 2
6 2
7 3
8 3
9 Manual_Mode_1
10 Manual_Mode_1
11 4
12 4
dtype: object
I have code that iterates over this series and produces the correct result when the series is passed to my object. The setter is:
#step_series.setter
def step_series(self, ss):
"""
On assignment, give the manual mode steps a unique name. Leave
the steps done on recipe the same.
"""
manual_mode = "MAN.OP."
new_manual_mode_text = "Manual_Mode_{}"
counter = 0
continuous = False
for i in ss.index:
if continuous and ss.at[i] != manual_mode:
continuous = False
counter += 1
elif not continuous and ss.at[i] == manual_mode:
continuous = True
ss.at[i] = new_manual_mode_text.format(str(counter))
elif continuous and ss.at[i] == manual_mode:
ss.at[i] = new_manual_mode_text.format(str(counter))
self._step_series = ss
but this iterates over the entire dataframe and is the slowest part of my code other than reading the logfile over the network.
How can I detect these non-unique sections and rename them uniquely without iterating over the entire series? The series is a column selection from a larger dataframe so adding extra columns is fine if needed.
For the completed answer I ended up with:
#step_series.setter
def step_series(self, ss):
pd.options.mode.chained_assignment = None
manual_mode = "MAN.OP."
new_manual_mode_text = "Manual_Mode_{}"
newManOp = (ss=='MAN.OP.') & (ss != ss.shift())
ss[ss == 'MAN.OP.'] = 'Manual_Mode_' + (newManOp.cumsum()-1).astype(str)
self._step_series = ss
Here's one way:
steps = [1,2,2,'MAN.OP.','MAN.OP.',2,2,3,3,'MAN.OP.','MAN.OP.',4,4]
steps = pd.Series(steps)
newManOp = (steps=='MAN.OP.') & (steps != steps.shift())
steps[steps=='MAN.OP.'] += seq.cumsum().astype(str)
>>> steps
0 1
1 2
2 2
3 MAN.OP.1
4 MAN.OP.1
5 2
6 2
7 3
8 3
9 MAN.OP.2
10 MAN.OP.2
11 4
12 4
dtype: object
To get the exact format you listed (starting from zero instead of one, and changing from "MAN.OP." to "Manual_mode_"), just tweak the last line:
steps[steps=='MAN.OP.'] = 'Manual_Mode_' + (seq.cumsum()-1).astype(str)
>>> steps
0 1
1 2
2 2
3 Manual_Mode_0
4 Manual_Mode_0
5 2
6 2
7 3
8 3
9 Manual_Mode_1
10 Manual_Mode_1
11 4
12 4
dtype: object
There a pandas enhancement request for contiguous groupby, which would make this type of task simpler.
There is s function in matplotlib that takes a boolean array and returns a list of (start, end) pairs. Each pair represents a contiguous region where the input is True.
import matplotlib.mlab as mlab
regions = mlab.contiguous_regions(ser_orig == manual_mode)
for i, (start, end) in enumerate(regions):
ser_orig[start:end] = new_manual_mode_text.format(i)
ser_orig
0 1
1 2
2 2
3 Manual_Mode_0
4 Manual_Mode_0
5 2
6 2
7 3
8 3
9 Manual_Mode_1
10 Manual_Mode_1
11 4
12 4
dtype: object
I have two data frames as below:
Sample_name C14-Cer C16-Cer C18-Cer C18:1-Cer C20-Cer
0 1 1 0.161456 0.033139 0.991840 2.111023 0.846197
1 1 10 0.636140 1.024235 36.333741 16.074662 3.142135
2 1 13 0.605840 0.034337 2.085061 2.125908 0.069698
3 1 14 0.038481 0.152382 4.608259 4.960007 0.162162
4 1 5 0.035628 0.087637 1.397457 0.768467 0.052605
5 1 6 0.114375 0.020196 0.220193 7.662065 0.077727
Sample_name C14-Cer C16-Cer C18-Cer C18:1-Cer C20-Cer
0 1 1 0.305224 0.542488 66.428382 73.615079 10.342252
1 1 10 0.814696 1.246165 73.802644 58.064363 11.179206
2 1 13 0.556437 0.517383 50.555948 51.913547 9.412299
3 1 14 0.314058 1.148754 56.165767 61.261950 9.142128
4 1 5 0.499129 0.460813 40.182454 41.770906 8.263437
5 1 6 0.300203 0.784065 47.359506 52.841821 9.833513
I want to divide the numerical values in the selected cells of the first by the second and I am using the following code:
df1_int.loc[:,'C14-Cer':].div(df2.loc[:,'C14-Cer':])
However, this way I lose the information from the column "Sample_name".
C14-Cer C16-Cer C18-Cer C18:1-Cer C20-Cer
0 0.528977 0.061088 0.014931 0.028677 0.081819
1 0.780831 0.821909 0.492309 0.276842 0.281070
2 1.088785 0.066367 0.041243 0.040951 0.007405
3 0.122529 0.132650 0.082047 0.080964 0.017738
4 0.071381 0.190178 0.034778 0.018397 0.006366
5 0.380993 0.025759 0.004649 0.145000 0.007904
How can I perform the division while keeping the column "Sample_name" in the resulting dataframe?
You can selectively overwrite using loc, the same way that you're already performing the division:
df1_int.loc[:,'C14-Cer':] = df1_int.loc[:,'C14-Cer':].div(df2.loc[:,'C14-Cer':])
This preserves the sample_name col:
In [12]:
df.loc[:,'C14-Cer':] = df.loc[:,'C14-Cer':].div(df1.loc[:,'C14-Cer':])
df
Out[12]:
Sample_name C14-Cer C16-Cer C18-Cer C18:1-Cer C20-Cer
index
0 1 1 0.528975 0.061087 0.014931 0.028677 0.081819
1 1 10 0.780831 0.821910 0.492309 0.276842 0.281070
2 1 13 1.088785 0.066367 0.041243 0.040951 0.007405
3 1 14 0.122528 0.132650 0.082047 0.080964 0.017738
4 1 5 0.071380 0.190179 0.034778 0.018397 0.006366
5 1 6 0.380992 0.025758 0.004649 0.145000 0.007904