Question

我几乎要写完一个程序，该程序将迭代由两个csv文件组成的csv文件。我被困在最后一列中，该列假定将Damage_done> 700000列为“高”，Damage_done <列为“低”，300000 <= Damage_done <= 699999列为“中”。我尝试编写一个循环并直接分配，但是抛出了以下错误：TypeError：（“ str”和'int'“的实例之间不支持”“>”，“发生在索引0'）。

1。清晰度质量（行）：如果（row ['damage_done']> 700000）： df3 ['dps_quality'] ='高' 如果（row ['damage_done'] <300000）： df3 ['dps_quality'] ='低' 如果（300000 <= row ['damage_done'] <= 699999）： df3 ['dps_quality'] ='中等'

df3['dps_quality'] = df3.apply(quality, axis = 1)
df3

和 2。

df3['dps_quality'][df3['damage_done'] > 700000] = 'High'
df3['dps_quality'][df3['damage_done'] < 300000] = 'Low'
df3['dps_quality'][300000 <= df3['damage_done'] <= 699000] = 'High'

import pandas as pd
import io
import requests as r

url = 'http://drd.ba.ttu.edu/isqs6339/hw/hw2/'
path = '/Users/jeredwilloughby/Desktop/Business Intelligence/'
file1 = 'players.csv'
file2 = 'player_sessions.csv'
fileout = 'pandashw.csv'

res1 = r.get(url + file1)
res1.status_code
df1 = pd.read_csv(io.StringIO(res1.text), delimiter='|')
df1

res2 = r.get(url + file2)
res2.status_code
df2 = pd.read_csv(io.StringIO(res2.text), delimiter=',')
df2.head(5)
df2.tail(5)

df3 = df1.merge(df2, how="left", on="playerid")
df3.describe()
list(df3)
df3.count()

df3['damage_done'].fillna(0, inplace=True)
df3.count()

df3.to_csv(path + fileout)

def performance(row):
    return (row['damage_done']*2.5 + row['healing_done']*4.5)/4

df3['player_performance_metric'] = df3.apply(performance, axis = 1)
df3
df3.to_csv(path + fileout)

def quality(row):
    if (row['damage_done'] > 700000):
        df3['dps_quality'] = 'High'
    if (row['damage_done'] < 300000):
        df3['dps_quality'] = 'Low'
    if (300000 <= row['damage_done'] <= 699999):
        df3['dps_quality'] = 'Medium'

df3['dps_quality'] = df3.apply(quality, axis = 1)
df3

预期：cvs文件输出将有一个标题为'dps_quality'的新列，并带有相关的高，中，低值。

实际：TypeError ：（在'str'和'int'“，“发生在索引0'的实例之间不支持”“>）。

Answer 1

列damage_done应当包含数字对象（int或float），而不是strings。
方法.apply为每一行调用函数quality。
他的函数返回的值将构成该方法将返回的序列。
如代码中所写，此系列将分配到数据框中的列dps_quality。
因此，无需在函数内使用列名。

考虑到这两个，可能的解决方法是：

def quality(damage_done):
    # this line assures that the value will be interpreted as an integer
    damage_done = int(damage_done)
    if damage_done > 700000:
        # now we are returning a value, instead of assigning it directly to the column
        return 'High'
    if damage_done < 300000:
        return 'Low'
    # removing the last check as it is not necessary
    return 'Medium'

# we are using the .apply method only on a series. This makes the reading easier
df3['dps_quality'] = df3['damage_done'].apply(quality)

用<，>和<= x <=

1 个答案: