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我一直在尝试操作一些代码来查看第 3 列中的不同uniqueclassindicator，如果第 3 列是2然后查看同一行中的前一行uniqueclassindicator以从该行检索数据。此外，我的输出只有在第 3 列和第 6 列都是1.

我一直在使用的代码：

from collections import defaultdict
import csv

# you probably can think up better names
fields = ('TitleA', 'TitleB', 'TitleIndicator', 'TitleRNum', 'TitleC', 'TitleD', 'TitlePNum', 'TitleBF', 'TitleCheck')

entries = defaultdict(dict)

with open("exampledata.csv", 'rb') as fd:
    reader = csv.DictReader(fd, fields)

    for counter, row in enumerate(reader):
        if counter != 0:
            TitleRNum = int(row['TitleRNum'])

            if row['TitlePNum']:
                TitlePNum = int(row['TitlePNum'])
            else:
                TitlePNum = ""

            Check = row['TitleCheck']
            Name = row['TitleB']

            key = (TitleRNum, TitleRNum)
            previous = entries[row['TitleIndicator']]

            if Check:
                # Scenario 1
                if (1, 1) in previous:
                    if (key[0] == 2 and key[1]>=2) or key[1] is None: # If Rank 2 and Position is Anything


                    if TitleRNum == 2:
                        p = previous[(2, 1)]
                        print '{p[TitleB]} {r[TitleB]} {p[TitleRNum]} {r[TitleRNum]} {p[TitlePNum]} {r[TitlePNum]} {p[TitleBF]} {r[TitleBF]} {p[TitleCheck]} {r[TitleCheck]}'.format(p=p, r=row)

            # remember this row for later rows to match against.
            previous[key] = row

示例数据：

TitleA,TitleB,TitleIndicator,TitleRNum,TitleC,TitleD,TitlePNum,TitleBF,TitleCheck
DataA,Joe,uniqueclassindicator1,1,125,3.659819202,1,984.2,Yes
DataA,Bob,uniqueclassindicator1,2,125,4.212490883,2,994.2,Yes
DataA,Dilon,uniqueclassindicator1,3,125,10.4587985,3,1023.2,Yes
DataA,Jessie,uniqueclassindicator1,4,125,12.68794408,9,,
DataA,Sammy,uniqueclassindicator1,5,125,13.91781792,9,,
DataA,Tommy,uniqueclassindicator1,6,125,15.68662715,9,,
DataA,Bobby,uniqueclassindicator1,7,125,16.45724048,9,,
DataA,Liam,uniqueclassindicator1,8,125,17.80601624,9,,
DataA,George,uniqueclassindicator1,9,125,29.77286312,9,,
DataA,Aaron,uniqueclassindicator1,10,125,34.68115514,9,,
DataA,Tim,uniqueclassindicator2,1,125,4.495485874,3,992.13,E
DataA,Tom,uniqueclassindicator2,2,125,4.899056741,2,951.1,E
DataA,Ryan,uniqueclassindicator2,3,125,6.856404461,1,932.1,E
DataA,Jack,uniqueclassindicator2,4,125,8.773751853,9,,
DataA,Jennifer,uniqueclassindicator2,5,125,9.16233388,9,,
DataA,Sarah,uniqueclassindicator2,6,125,10.16514897,9,,
DataA,Joanne,uniqueclassindicator2,7,125,18.43432845,9,,
DataA,Jess,uniqueclassindicator2,8,125,19.28867787,9,,
DataA,Test1,uniqueclassindicator3,1,125,4.634033328,9,,
DataA,Test2,uniqueclassindicator3,2,125,6.470141577,9,,
DataA,Test3,uniqueclassindicator3,3,125,7.337664044,3,33.59,Yes
DataA,Test4,uniqueclassindicator3,4,125,7.614649866,1,30,D
DataA,Test5,uniqueclassindicator3,5,125,10.02454642,9,,
DataA,Test6,uniqueclassindicator3,6,125,11.95552344,9,,
DataA,Test7,uniqueclassindicator3,7,125,14.36021059,2,33.58,E
DataA,Test8,uniqueclassindicator3,8,125,14.73472778,9,,
DataA,Test9,uniqueclassindicator3,9,125,24.30844993,9,,
DataA,Mark,uniqueclassindicator4,1,125,4.634033328,1,395.1,F
DataA,Jason,uniqueclassindicator4,2,125,6.470141577,9,,
DataA,Dan,uniqueclassindicator4,3,125,7.337664044,9,,
DataA,Tessa,uniqueclassindicator4,4,125,7.614649866,2,395.2,F
DataA,Laura,uniqueclassindicator4,5,125,10.02454642,9,,
DataA,Lizz,uniqueclassindicator4,6,125,11.95552344,9,,
DataA,Hannah,uniqueclassindicator4,7,125,14.36021059,3,395.3,F
DataA,Ian,uniqueclassindicator4,8,125,14.73472778,9,,
DataA,Janet,uniqueclassindicator4,9,125,24.30844993,9,,

期望的输出：

Joe,Bob,1,2,1,2,984.2,994.2,Yes,Yes
Mark,Jason,1,2,1,9,295.1,,F,,

因此，为了澄清 TitleIndicator/Uniqueclassindicator 定义的每个组，如果第 6 列和第 3 列都等于 1，我希望能够从该组的前两行中提取数据。

如果有人能告诉我如何修复此代码，将不胜感激。非常感谢

4

3 回答 3

5

好吧，比赛结束了，但我仍然想给出我的解决方案。以下是详细的评论答案：

# Import "csv.DictReader" and put it in the name "dr".
from csv import DictReader as dr

# These are the columns we will be working with.
cols = "TitleB", "TitleRNum", "TitlePNum", "TitleBF", "TitleCheck"

# This is a variable to hold a previous row for future processing.
# It severs the same purpose as the "entries" dict in Sudipta Chatterjee's answer.
# I set it to 0 simply so its value is false.  You could also set it to "False" or "None".
mark = 0

# Open the CSV file in binary mode.
with open("exampledata.csv", "rb") as f:

    # This loops through what is returned by "DictReader".
    #
    # The expression "f.readline().strip().split(",")" reads the first line of the file,
    # (which is the column names), strips off the newline at the end,
    # and then gets the column names by splitting the line on commas.
    for row in dr(f, f.readline().strip().split(",")):

        # This checks if "mark" is true.
        # If it is, then that means "mark" contains a previous row to be processed.
        if mark:

            # This line takes the row stored in "mark" as well as the current row
            # and puts them together, separating the values with commas using "str.join".
            print ",".join([",".join([mark[c], row[c]]) for c in cols])

        # This is a compact statement equivalent to:
        #
        #    if row["TitlePNum"] == row["TitleRNum"] == "1":
        #        mark = row
        #    else:
        #        mark = 0
        #
        # It sees if the "TitlePNum" and "TitleRNum" columns in the current row are both "1".
        # If so, it saves that row in "mark" for future processing.
        #
        # It is basically the same thing as the
        #
        #    if TitleRNum == 1 and TitlePNum == 1:
        #        entries[row['TitleIndicator']] = row
        #
        # part in Sudipta Chatterjee's answer.
        mark = row if row["TitlePNum"]==row["TitleRNum"]=="1" else 0

以下是正常情况下的答案：

from csv import DictReader as dr
cols = "TitleB", "TitleRNum", "TitlePNum", "TitleBF", "TitleCheck"
mark = 0
with open("exampledata.csv", "rb") as f:
    for row in dr(f, f.readline().strip().split(",")):
        if mark: 
            print ",".join([",".join([mark[c], row[c]]) for c in cols])
        mark = row if row["TitlePNum"]==row["TitleRNum"]=="1" else 0

输出：

Joe,Bob,1,2,1,2,984.2,994.2,Yes,Yes
Mark,Jason,1,2,1,9,395.1,,F,

如您所见，我的解决方案更小更高效。

于 2013-10-27T17:01:10.477 回答
2

这有效：

from collections import defaultdict
import csv

# you probably can think up better names
fields = ('TitleA', 'TitleB', 'TitleIndicator', 'TitleRNum', 'TitleC', 'TitleD', 'TitlePNum', 'TitleBF', 'TitleCheck')

entries = defaultdict(dict)

with open("exampledata.csv", 'rb') as fd:
    reader = csv.DictReader(fd, fields)

    for counter, row in enumerate(reader):
        if counter != 0:
            TitleRNum = int(row['TitleRNum'])

            # If this row has a TitlePNum, keep it, otherwise reset to -1
            TitlePNum = -1
            if row['TitlePNum']:
                TitlePNum = int(row['TitlePNum'])

            # If we have already seen a row with the same class 
            # that has 1 at both RNum and PNum,
            # use that to print locally
            if row['TitleIndicator'] in entries:
                previousRow = entries[row['TitleIndicator']]
                currentRow = row

                itemsToPrint = ['TitleB', 'TitleRNum', 'TitlePNum', 'TitleBF', 'TitleCheck']
                output = ""
                for item in itemsToPrint:
                    output += previousRow[item] + ',' + currentRow[item] + ','

                # Finally, strip the last comma and print
                output = output[:-1]
                print output

                # Remove the previous entry from the dict
                del entries[row['TitleIndicator']]


            # If both RNum and PNum are 1, then save this as a candidate for future reference
            if TitleRNum == 1 and TitlePNum == 1:
                entries[row['TitleIndicator']] = row
于 2013-10-24T01:00:27.887 回答
2

您在帖子的标题中使用了“键”——所以我在这里提供了一个带有字典的解决方案 :) 哇（！）能够做到这一点感觉很好。

from csv import DictReader

# 1) read in the data and store it row-wise in the list 'data'
data, numclasses = [], []
with open("exampledata.csv", 'rb') as fd:
    reader = DictReader(fd)
    for counter, row in enumerate(reader):
      data.append(row)
      numclasses.append(row['TitleIndicator'][-1])
numclasses = len(list(set(numclasses))) # returns unique no. of classes

# 2) group data in a dictionary where each key uniquely corresponds to a class
datagrouped = {"class%s"%(i + 1): [] for i in range(numclasses)}
for row in data:
  classID = row['TitleIndicator'][-1]
  datagrouped["class%s"%classID].append(row)

# 3) go through each class within the dictionary, then go through the data
# within the class (row-wise), and print out rows that meet requirements.
for classname in datagrouped.keys(): # class loop
  uniq_class = datagrouped[classname]
  for i, row in enumerate(uniq_class): # row loop
    if i > 0:
      cond1 = row['TitleRNum'] == '2'
      prev_row = uniq_class[i - 1]
      cond2 = prev_row['TitleRNum'] == '1' and prev_row['TitlePNum'] == '1'
      if cond1 & cond2:
        print ["%s"%x for x in prev_row.itervalues()]
        print ["%s"%x for x in row.itervalues()]

当我exampledata.csv在同一目录中运行它时，我得到以下输出：

['DataA', 'Mark', '125', '4.634033328', '1', '395.1', '1', 'F', 'uniqueclassindicator4']
['DataA', 'Jason', '125', '6.470141577', '2', '', '9', '', 'uniqueclassindicator4']
['DataA', 'Joe', '125', '3.659819202', '1', '984.2', '1', 'Yes', 'uniqueclassindicator1']
['DataA', 'Bob', '125', '4.212490883', '2', '994.2', '2', 'Yes', 'uniqueclassindicator1']
于 2013-10-27T13:17:59.357 回答